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2 years ago

8

I need help with number 3

2 answers:

Julli [10]2 years ago

7 0

The answer is A) 2.25 each point goes by 0.10 and it's so in between 2.20 and 2.30

Illusion [34]2 years ago

5 0

A. Hope this helps!!!

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MATH HELP!!!!!<br><br> If y = 15 when x = 2, what is the value of x when y = 7?

Ede4ka [16]

<u>Direct Variation:</u> $\frac{y}{x} = k$

$\frac{15}{2} = k$

when y = 7: $\frac{7}{x} = \frac{15}{2}$

14 = 15x

$\frac{14}{15} = x$

<u>Inverse variation:</u> y*x = k

15 * 2 = k

30 = k

when y = 7: 7 * x = 30

$x = \frac{30}{7}$

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2 years ago

Which of the following are square roots of the number below 529?

exis [7]

<span>-10
-(100) ^1/2
</span><span>100 ^1/2 </span>

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2 years ago

I need help with this question! Will mark brainliest :)

PolarNik [594]

C is 3/5. D is 1. B is underlined

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2 years ago

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Use the intermediate value theorem to find the value of c such that f(c) = M. f(x) = x^2 - x + 1 text( on ) [1,8]; M = 21 c =

DanielleElmas [232]

Answer:

$c = 5$

Step-by-step explanation:

Given

$f(c) = M$

$f(x) = x^2 - x + 1$

$Interval: [1,8]$

$M = 21$

Required

Find c using Intermediate Value theorem

First, check if the value of M is within the given range:

$f(x) = x^2 - x + 1$

$f(1) = 1^2 - 1 + 1$

$f(1) = 1$

$f(x) = 8^2 - 8 + 1$

$f(x) = 57$

$1 \le M \le 57$

$1 \le 21 \le 57$

M is within range.

Solving further:

We have:

$f(c) = f(x) = M$

$f(x) = 21$

Substitute 21 for f(x) in $f(x) = x^2 - x + 1$

$21 = x^2 - x + 1$

Express as quadratic function

$x^2 - x + 1 - 21 = 0$

$x^2 - x - 20 = 0$

Expand

$x^2 + 4x - 5x - 20$

$x(x+4)-5(x+4)=0$

$(x - 5)(x+4) = 0$

$x - 5 = 0$ or $x + 4= 0$

$x = 5$ or $x = -4$

The value of $x = -4$ is outside the $Interval: [1,8]$

So:

$x = 5$

$f(c) = f(x) = M$

$f(c) = f(5) = 21$

By comparison:

$c = 5$

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2 years ago

The weight of an elephant is 1000 times the weight of a cat. If the elephant weighs 14,000 pounds how many pounds does the cat w

ohaa [14]

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