Solve the following system of equations.
(15 Points)
- 2x + y = -20 & 4y + 3x = 30
PLEASE HELP ME

Answer:

force = 100 m/s/25 s = 4 m/s^2

Answer:

I think it B or C but hope I can help :)

Step-by-step explanation:

idkStep-by-step explanation:

Answer:

21 quarters and 19 dimes

Step-by-step explanation:

Create a system of equations where q is the number of quarters and d is the number of dimes.

0.25q + 0.1d = 7.15

q + d = 40

Solve by elimination by multiplying the bottom equation by -0.25

0.25q + 0.1d = 7.15

-0.25q - 0.25d = -10

Add them together and solve for d:

-0.15d = -2.85

d = 19

Then, plug in 19 as d into the second equation to solve for q:

q + d = 40

q + 19 = 40

q = 21

So, there are 21 quarters and 19 dimes

Answer:

21 Quarters, 19 Dimes.

Step-by-step explanation:

21 x .25 = 5.25

19 x .10 = 1.90

5.25 + 1.90 = 7.15

Answer:

0 degrees faranhiet

Step-by-step explanation:

when you add 14 and -5 you get 9 then the question says 9 degrees cooler so that would translate to the problem 9 - 9 which you get zero.

Answer:

- 1/2

Step-by-step explanation:

Just took the test

The value of cot (b) using trigonometry is -1/2 when tan b = -2.

Since we are given that tan(b) = -2 and the terminal side of angle b is in quadrant II, determine the value of cot(b).

In quadrant II, the sine and cosine values are positive, while the tangent and cotangent values are negative.

Given:

tan(b) = -2

The cotangent (cot) of an angle is the reciprocal of the tangent:

cot(b) = 1 / tan(b)

Now, substituting the given value of tan(b):

cot(b) = 1 / (-2)

cot(b) = -1/2

So, the value of cot(b) is -1/2.

Learn more about Trigonometry here:

https://brainly.com/question/11016599

#SPJ4

Answer: 1

Step-by-step explanation: 863x14ysqrt9= 3 simplified is 1

Answer:

- 8

Step-by-step explanation:

Step 1:

12x ÷ 4y         Equation

Step 2:

12 ( - 8 ) ÷ 4 ( 3 )     Input x and y

Step 3:

- 96 ÷ 12      Multiply

Answer:

- 8      Divide

Hope This Helps :)

Answer:

0 The product will be smaller than both the factors.

0 There will be a zero in the tenths place.

0 The answer is 0.04.

Answer: $41.472 or $41.50

Step by step solving: First convert 20% to a decimal which is .20 now multiply that by $51.84 which equals 10.368 now subtract that number from 51.84 which equals 41.472

If you are allowed to round to the nearest hundreth then that would be $41.50 before the tip.

Answer:

Yess

Step-by-step explanation:

If you multiply all the numbers together you get 39690000. And the square root of that number is 6300. So it is a perfect square. If you did 6300² = 39690000.

Answer:

x= 6, -7

Step-by-step explanation:

Answer:

1/6

Step-by-step explanation:

9: 9,18,27,36,45,54,63,72,81,90

10: 10,20,30,40,50,60,70,80,90,100

Since, 90 is the LCM of 9 and 10 that turns into your denominater then your equation turns into this...

5/90 x 3/90

The denomineter stays the same so you get 90 then 5 x 3 equals 15 so you get 15 and in total that gives you 15/90. Divide the top and bottom by the greatest number that will divide both numbers exactly which gives you 1/6

HOPE THIS HELPED :D

Answer:

B. 61

Step-by-step explanation:

Given:

∆PQR ≅ ∆PQS

PQ = 2x + 41

QS = 7x - 24

QR = 3x + 16

Required:

Numerical value of PQ

SOLUTION:

First, create an equation to find the value of x as follows:

Since both triangles are congruent, therefore:

QS = QR

7x - 24 = 3x + 16 (Substitution)

Collect like terms

7x - 3x = 24 + 16

4x = 40

Divide both sides by 4

4x/4 = 40/4

x = 10

Find PQ by plugging x = 10 into PQ = 2x + 41

PQ = 2(10) + 41

PQ = 20 + 41

PQ = 61

Answer:

-1.0

Step-by-step explanation:

Answer:

7

Step-by-step explanation:

Step 1: translate the given word problem into solvable algebraic equation.

Let "t" represent the number of tadpoles Joseph caught.

"Wendy caught nine times three more than Joseph caught" can be expressed as 9(t + 3)

Number of Wendy's tadpole = [tex] 9(t + 3) [/tex]

Given that half of Wendy's tadpole and also 6 times Joseph's tadpoles swarm away leaving Wendy with only 3, the following equation can be written:

[tex] \frac{9(t + 3)}{2} - 6t = 3 [/tex]

Step 2: solve for t using the equation created.

[tex] \frac{9(t + 3)}{2} - 6t = 3 [/tex]

Add 6t to both sides

[tex] \frac{9(t + 3)}{2} - 6t + 6t = 3 + 6t [/tex]

[tex] \frac{9(t + 3)}{2} = 3 + 6t [/tex]

Multiply both sides by 2

[tex] \frac{9(t + 3)}{2}*2 = (3 + 6t)*2 [/tex]

[tex] 9(t + 3) = 2(3 + 6t) [/tex]

[tex] 9t + 27 = 6 + 12t [/tex]

Collect like terms

[tex] 9t - 12t = 6 - 27 [/tex]

[tex] -3t = -21 [/tex]

Divide both sides by -3

[tex] t = 7 [/tex]

Joseph caught 7 tadpoles

Answer:

a. 3

Step-by-step explanation:

Answer:

(A)  X = 3

Step-by-step explanation:

-_-

Answer:

C.1%

Step-by-step explanation:

Answer:

[tex]m = \frac{2}{3}[/tex]

Step-by-step explanation:

Use the following equation:

slope [tex](m) = \frac{y^2 - y^1}{x^2 - x^1}[/tex]

Set the variables with the points:

[tex](x_1 , y_1) = (-16 , -6)\\(x_2 , y_2) = (-4 , 2)\\[/tex]

Plug in the corresponding numbers to the corresponding variables in the equation:

[tex]m = \frac{2 - (-6)}{-4 - (-16)}[/tex]

[tex]m = \frac{2 - (-6)}{- 4 - (-16)} = \frac{2 + 6}{16 - 4} = \frac{8}{12}[/tex]

slope [tex]m = \frac{8}{12}[/tex] [tex]= \frac{2}{3}[/tex]  (simplified)

~

Answer:

2/3

Step-by-step explanation:

Answer:

[tex] {a}^{2} + 2ab + {b}^{2} [/tex]

Answer:

(a + b) × (a + b)

Step-by-step explanation:

(a + b)²

(a + b) × (a + b)

Answer:

9-4/5

5/5 = 1

Step-by-step explanation:

Answer:

1

9-4/5 is equal to 5/5 which is equal to 1

Answer:

5=148 and 8=32

Step-by-step explanation:

For this equation we know since the lines are parelle that they will produce the same angle degrees.

so 2,4,6,and 8 are all 32

if we subtract 180 (the dregree of a stright line) from 32 we get 148

so 148 is for 1,3,5, and 7

Answer:

Step-by-step explanation:

Area of rectangle= Length x Width

and the given Area is a quadratic expression

A=[tex]x^{2} -4x-21[/tex]

We use the factorization method so,

we need 2 numbers that when multiplied we get -21 and when we add/subtract we get -4 so,

A=[tex]x^{2} -7x+3x-21[/tex]

now we simplify,

A=[tex]x(x-7)+3(x-7)[/tex]

A=[tex](x-7)(x+3)[/tex]

This looks familiar doesn't it, when we write the formula for the area of rectangle its

A= Length x Width and the equation here shows that

A= [tex](x-7)(x+3)[/tex]

So the expression for the length is x-7 and

the expression for the width is x+3 i think u have missed maybe some information on the question as such that the perimeter might be missing because length could be either x-7 or even x+3 same goes for width maybe someone can correct me if im wrong

Answer:

i need a picture

Step-by-step explanation:

Answer:

Option (A)

Step-by-step explanation:

Given expression is,

[tex]5\sqrt{48a}=60[/tex]

Squaring on both the sides of the equation,

[tex](5\sqrt{48a})^2=(60)^2[/tex]

25(48a) = 3600

1200a = 3600

By dividing the equation by 1200,

a = [tex]\frac{3600}{1200}[/tex]

a = 3

Therefore, Option (A) will be the answer.

answer:

48p² + 4p - 4

solution:

8p × 6p +8p × 2 - 2 × 6p - 2 × 2

48p² + 8p x 2 - 2 ×6p - 2 × 2

48p² + 16p - 12p - 4

48p² + 4p - 4

The required algebraic product is 48[tex]p^{2}[/tex] + 4p - 4.

Given the two binomial (8p - 2)(6p + 2).

To multiply two binomials, take the first term of the first binomial and multiply with the entire second binomial and positive or negative as per given in the first binomial ,take the second term of the first binomial and multiply with the entire second binomial.

Let a, b, c and d be any four variables. Consider (a - b)(c + d) gives

a(c + d) - (c + d).

That implies, (8p - 2)(6p + 2) = 8p(6p + 2) - 2(6p + 2)

Multiply by removing the brackets gives,

(8p - 2)(6p + 2) = 48[tex]p^{2}[/tex] + 16p - 12p - 4

Combining like terms and algebraic sum gives,

(8p - 2)(6p + 2) = 48[tex]p^{2}[/tex] + 4p - 4.

Hence, the required algebraic product is 48[tex]p^{2}[/tex] + 4p - 4.

Learn more about binomials, click here:

https://brainly.com/question/29164885

#SPJ6

Answer:

11, there both 8 away so its right in ze middle :D

Step-by-step explanation:

Using the given definition, for [tex]f(x)=\frac1{\sqrt x}+x[/tex], we have

[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{\left(\frac1{\sqrt{x+h}}+x+h\right)-\left(\frac1{\sqrt x}+x\right)}h[/tex]

Right away, we see x and -x in the numerator, so we can drop those terms.

[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{\frac1{\sqrt{x+h}}+h-\frac1{\sqrt x}}h[/tex]

Remember that limits distribute over sums, i.e.

[tex]\displaystyle\lim_{x\to c}(f(x)+g(x))=\lim_{x\to c}f(x)+\lim_{x\to c}g(x)[/tex]

so we can separate the h from everything else in the numerator:

[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{\frac1{\sqrt{x+h}}-\frac1{\sqrt x}}h+\lim_{h\to0}\frac hh[/tex]

Since h ≠ 0, we have [tex]\frac hh=1[/tex], so the second limit is simply 1.

[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{\frac1{\sqrt{x+h}}-\frac1{\sqrt x}}h+1[/tex]

For the remaining limit, focus on the numerator for now. Combine the fractions in the numerator:

[tex]\dfrac1{\sqrt{x+h}}-\dfrac1{\sqrt x}=\dfrac{\sqrt x-\sqrt{x+h}}{\sqrt x\sqrt{x+h}}[/tex]

Recall the difference of squares identity,

[tex]a^2-b^2=(a-b)(a+b)[/tex]

Let [tex]a=\sqrt x[/tex] and [tex]b=\sqrt{x+h}[/tex]. Multiply the numerator and denominator by [tex](a+b)[/tex], so that the numerator can be condensed using the identity above.

[tex]\dfrac{\sqrt x-\sqrt{x+h}}{\sqrt x\sqrt{x+h}}\cdot\dfrac{\sqrt x+\sqrt{x+h}}{\sqrt x+\sqrt{x+h}}[/tex]

[tex]=\dfrac{(\sqrt x)^2-(\sqrt{x+h})^2}{\sqrt x\sqrt{x+h}(\sqrt x+\sqrt{x+h})}[/tex]

[tex]=\dfrac{x-(x+h)}{\sqrt x\sqrt{x+h}(\sqrt x+\sqrt{x+h})}[/tex]

[tex]=-\dfrac h{\sqrt x\sqrt{x+h}(\sqrt x+\sqrt{x+h})}[/tex]

Back to the limit: all this rewriting tells us that

[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{-\frac h{\sqrt x\sqrt{x+h}(\sqrt x+\sqrt{x+h})}}h+1[/tex]

Again, the h's cancel, and we can pull out the factor of -1 from the numerator and simplify the fraction:

[tex]f'(x)=\displaystyle-\lim_{h\to0}\frac1{\sqrt x\sqrt{x+h}(\sqrt x+\sqrt{x+h})}+1[/tex]

The remaining expression is continuous at h = 0, so we can evaluate the limit by substituting directly:

[tex]f'(x)=-\dfrac1{\sqrt x\sqrt{x+0}(\sqrt x+\sqrt{x+0})}+1[/tex]

[tex]f'(x)=-\dfrac1{2x\sqrt x}+1[/tex]

or, if we write [tex]\sqrt x=x^{1/2}[/tex], we get

[tex]f'(x)=-\dfrac12x^{-3/2}+1[/tex]