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2 years ago

Evaluate p^2 - r^2 for p = 10, r = -4 and t = 5. 116 92 84

Mathematics

1 answer:

eduard2 years ago

6 0

P² - r²

subtitute p =10 and r= -4

10² - (-4)² = 100 - 16 = 84

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A. seven tents b. seven hundredths c. seven tents

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2 years ago

Identify a pattern in the given list of numbers. Then use this pattern to find the next number. 17,7,-3,-13,-23

notka56 [123]

Answer:

-33

Step-by-step explanation:

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1 year ago

How can I group these numbers 15 divide 3x5+4x8 to equal 40

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15/3x5+4x8=40
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2 years ago

Read 2 more answers

Julie and Eric row their boat (at a constant speed) downstream for 40 miles in 4 hours, helped by the current. Rowing at the sam

Liula [17]

Answer: 3 miles per hour

Step-by-step explanation:

Use the formula "distance (d) = rate (r) x time (t)" to create a system of equations.

Let "r" represent the rate they are rowing

Let "c" represent the current

time rate distance EQUATION

Downstream: 4 hours r + c 40 miles 4(r + c) = 40

Upstream: 10 hours r - c 40 miles 10(r - c) = 40

Distribute, then eliminate r to solve for c:

Down: 4r + 4c = 40 → 5(4r + 4c = 40) → 20r + 20c = 200

Up: 10r - 10c = 40 → -2(10r - 10c = 40) → -20r + 20c = -80

40c = 120

÷40 ÷40

c = 3

4 0

2 years ago

CALC- limits please show your method

gladu [14]

A. Factor the numerator as a difference of squares:

$\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6$

c. As $x\to\infty$ , the contribution of the terms of degree less than 2 becomes negligible, which means we can write

$\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4$

e. Let's first rewrite the root terms with rational exponents:

$\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}$

Next we rationalize the numerator and denominator. We do so by recalling

$(a-b)(a+b)=a^2-b^2$
$(a-b)(a^2+ab+b^2)=a^3-b^3$

In particular,

$(x^{1/3}-x)(x^{2/3}+x^{4/3}+x^2)=x-x^3$
$(x^{1/2}-x)(x^{1/2}+x)=x-x^2$

so we have

$\displaystyle\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}\cdot\frac{x^{2/3}+x^{4/3}+x^2}{x^{2/3}+x^{4/3}+x^2}\cdot\frac{x^{1/2}+x}{x^{1/2}+x}=\lim_{x\to1}\frac{x-x^3}{x-x^2}\cdot\frac{x^{1/2}+x}{x^{2/3}+x^{4/3}+x^2}$

For $x\neq0$ and $x\neq1$ , we can simplify the first term:

$\dfrac{x-x^3}{x-x^2}=\dfrac{x(1-x^2)}{x(1-x)}=\dfrac{x(1-x)(1+x)}{x(1-x)}=1+x$

So our limit becomes

$\displaystyle\lim_{x\to1}\frac{(1+x)(x^{1/2}+x)}{x^{2/3}+x^{4/3}+x^2}=\frac{(1+1)(1+1)}{1+1+1}=\frac43$

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2 years ago