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3 years ago

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The number of boys to girls in math class is 2:3 If there are 21 girls in Mrs. Vest's math class, how many boys are there? Can w

e use what we know about quivalent ratios to figure out how many boys are in the class?

2 answers:

slega [8]3 years ago

6 0

Answer:

There are 14 boys in the class. Yes, we can use what we know to figure out how many boys are in the class.

Step-by-step explanation:

think about the ratio as a fraction if the fraction is 2/3 and then compare it to x/21 21/3 is 7 and 7*2 is 14

torisob [31]3 years ago

5 0

Answer:

there are 6 boys in the class

Step-by-step explanation:

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If the speed of a is 3 ft/s up and is increasing at 4 what is the instantaneous speed and acceleration

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3 years ago

A person standing close to the edge on the top of a 400-ft building drops a baseball. The quadratic function h left parenthesis

swat32

Answer:

$h(0)=400$

The above function value for $h(0)$ shows that height of the ball before it was dropped i.e. at time = 0 seconds.

The height of the ball above the ground before it was dropped = 400 ft.

Step-by-step explanation:

Given:

The quadratic function that models the height of the baseball above the ground in feet , $t$ seconds after it was dropped is given as:

$h(t)=-16t^2+400$

To find $h(0)$ and interpret the meaning of the function value in contect of the problem.

Solution:

In order to find $h(0)$ , we will replace $t=0$ in the given function as $h(t)$ is a function of time $t$ .

Thus, we have:

$h(0)=-16(0)^2+400$

$h(0)=16(0)+400$

$h(0)=0+400$

$h(0)=400$ (Answer)

The above function value for $h(0)$ shows that height of the ball before it was dropped i.e. at time = 0 seconds.

The height of the ball above the ground before it was dropped = 400 ft.

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3 years ago

WILL MARK BRAINLIEST!!! 40 POINTS!! ACTUAL ANSWERS, PLZZZ

steposvetlana [31]

Answer:

Part A:

$\left(x + 7\right)^{5}=x^{5} + 35 x^{4} + 490 x^{3} + 3430 x^{2} + 12005 x + 16807$

Part B:

The closure property describes cases when mathematical operations are CLOSED. It means that if you apply certain mathematical operations in a polynomial it will still be a polynomial. Polynomials are closed for sum, subtraction, and multiplication.

It means:

$\text{Sum of polynomials } \Rightarrow \text{ It will always be a polynomial}$

$\text{Subtraction of polynomials } \Rightarrow \text{ It will always be a polynomial}$

$\text{Multiplication of polynomials } \Rightarrow \text{ It will always be a polynomial}$

But when it is about division:

$\text{Division of polynomials } \Rightarrow \text{ It will not always/sometimes be a polynomial}$

<u>Example of subtraction of polynomials:</u>

<u /> $(2x^2+2x+3) - (x^2+5x+2)$ <u />

<u /> $x^2-3x+1$ <u />

Step-by-step explanation:

<u>First, it is very important to define what is a polynomial in standard form: </u>

It is when the terms are ordered from the highest degree to the lowest degree.

Therefore I can give:

$x^5-5x^4+3x^3-3x^2+7x+20$

but,

$x^5+3x^3-3x^2+7x+20-5x^4$ is not in standard form.

For this question, I can simply give the answer: $x^5-5x^4+3x^3-3x^2+7x+20$ and it is correct.

But I will create a fifth-degree polynomial using this formula

$(a+b)^n=\sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$

Also, note that

$\binom{n}{k}=\frac{n!}{(n-k)!k!}$

For $a=x \text{ and } b=7$

$\left(x + 7\right)^{5}=\sum_{k=0}^{5} \binom{5}{k} \left(x\right)^{5-k} \left(7\right)^k$

$\text{Solving for } k \text{ values: } 0, 1, 2, 3, 4 \text{ and } 5$

Sorry but I will not type every step for each value of $k$

The first one is enough.

For $k=0$

$\binom{5}{0} \left(x\right)^{5-0} \left(7\right)^{0}=\frac{5!}{(5-0)! 0!}\left(x\right)^{5} \left(7\right)^{0}=\frac{5!}{5!} \cdot x^5= x^{5}$

Doing that for $k$ values:

$\left(x + 7\right)^{5}=x^{5} + 35 x^{4} + 490 x^{3} + 3430 x^{2} + 12005 x + 16807$

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Yuliya22 [10]

<span><span> x =(4-√-104)/6=(2-i√ 26 )/3= 0.6667-1.6997i</span><span> x =(4+√-104)/6=(2+i√ 26 )/3= 0.6667+1.6997i</span></span>

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3 years ago