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3 years ago

CORRECT ANSWER GETS BRAINLEYY

Mathematics

1 answer:

Ratling [72]3 years ago

5 0

Answer:

Step-by-step explanation:

$w = x + \frac{y}{z}\\\\w-x = \frac{y}{z}\\\\(w - x)*z = y\\\\z = \frac{y}{w-x}$

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Hamburgers cost $3 and hotdogs cost $2. The total cost of buying b hamburgers and d hotdogs can be expressed by C=3b+2d. How can

Temka [501]

C = 3b+2d is the same as 3b+2d = C

Let's isolate d. To do this, we first need to subtract 3b from both sides

3b+2d = C
3b+2d-3b = C-3b
2d = C-3b

Then divide both sides by 2

2d = C-3b
2d/2 = (C-3b)/2
d = (C-3b)/2

Take note of the parenthesis as they are very important. We want to divide ALL of C-3b over 2. We don't want to just divide -3b over 2.

The answer choices you have aren't 100% clear but I have a feeling your teacher meant to say d = (C-3b)/2 instead of d = C-3b/2 for choice A

If that assumption is correct, then the answer is choice A.

7 0

3 years ago

Find the highest common factor of 32 and 80

valina [46]

32=2*2*2*2*2=2^5
80=2*2*2*2*5

hcf = 2*2*2*2=2⁴=16

3 0

3 years ago

Read 2 more answers

The figure is formed from rectangles . Find the total area

Anna35 [415]

It might be c. im not really sure, but that's what I got. sorry if its wrong...

8 0

3 years ago

Solve 73 make sure to also define the limits in the parts a and b

Aleks04 [339]

73.

$f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}$

$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3$ $\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3$

Since we can't divide by zero, we need to find when:

$x^4-2x^2+144=0$

But before, we can factor the numerator and the denominator:

$\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}$

Now, we can conclude that the vertical asymptotes are located at:

$\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}$

so, for x = -3:

$\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty$ $\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty$

For x = 4:

$\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$ $\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$

4 0

1 year ago

What is the reciprocal of 17 1/2

Aneli [31]

Answer:

2/35

Step-by-step explanation:

If you convert 17 1/2 to a improper fraction, it becomes 35/2. Then to find the reciprocal, flip the numerator and denominator. So, 2/35

Hope this helps!

4 0

2 years ago