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3 years ago

What is 15 = -y + 10

2 answers:

7 0

It's a simple linear equation in a single variable, 'y' . Its solution is the number
that 'y' must be in order to make the equation a true statement. Here's how to
find the solution:

Write the equation: <span>15 = -y + 10

Subtract 10 from each side: 5 = -y

Multiply each side by -1 : <u>-5 = y</u>
</span>

gregori [183]3 years ago

7 0

15=-y+10
=-10+15=-y
=5=-y
=-5=y
welcome :)

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A student solving for the acceleration of an object has applied appropriate physics principles and obtained the expression a=a1+

egoroff_w [7]

Answer:

$a=\frac{3\times 7}{7}+\frac{12}{7}$ (First step for obtaining a common denominator for the two fraction)

$a=\frac{33}{7}\text{ m/s}^2$

Step-by-step explanation:

$\text{Given Expression: }a=a_1+\frac{F}{m}$

where,

a is acceleration of an object.(Need to calculate)

$a_1=3.00\text{ m/s}^2$

$F=12.0\text{ Kg.m/s}^2$

$m=7.00\text{ kg}$

$\text{Substitute }a_1, \text{ F, and m into expression and we get}$

$a=3+\frac{12}{7}$

Now we will simplify above expression for a

First we make common denominator.

Common denominator is 7. So, we make both denominator 7. We multiply by 7 at top and bottom with 3. We get

$a=\frac{3\times 7}{7}+\frac{12}{7}$ (First step for obtaining a common denominator for the two fraction)

Now we combine the numerator

$a=\frac{21+12}{7}$

$a=\frac{33}{7}\text{ m/s}^2$

3 0

3 years ago

Read 2 more answers

Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros.

butalik [34]

I'll try to understand this but I don't really get it because I'm in middle school

6 0

3 years ago

Find the sum. Write your answer in simplest form.

Masteriza [31]

Answer:

19/20

Step-by-step explanation:

write all numerators above the common denominator 20

14+5/20

19/20

4 0

3 years ago

Someone help me with my math homework pleaseee. Find the volumes of the pyramids and the height is 7cm for the first one.

gregori [183]

$\bf \textit{volume of a pyramid}\\\\
V=\cfrac{1}{3}Bh\qquad 
\begin{cases}
B=\textit{area of the base}\\
h=height
\end{cases}$

now, the first one, on the far-left.... can't see the height.. but I gather you do, now as far as its Base area, well, the bottom is just a 12x12 square, so the area of its base is just 12*12

now, the middle pyramid, has a height of 6, the base is also a square, 8x8, so the Base area is just 8*8

now the last one on the far-right

has a height of 8, the Base is a Hexagon, with sides of 6

$\bf \textit{area of a regular polygon}\\\\
A=\cfrac{1}{4}ns^2cot\left( \frac{180}{n} \right)\qquad 
\begin{cases}
n=\textit{number of sides}\\
s=\textit{length of one side}\\
\frac{180}{n}=\textit{angle in degrees}\\
----------\\
n=6\\
s=6
\end{cases}\\\\\\ A=\cfrac{1}{4}\cdot 6\cdot 6^2\cdot cot\left( \frac{180}{6} \right)$

5 0

2 years ago

Binomial Expansion/Pascal's triangle. Please help with all of number 5.

Mandarinka [93]

$\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}$

The rows add up to $1,2,4,8,16$ , respectively. (Notice they're all powers of 2)

The sum of the numbers in row $n$ is $2^{n-1}$ .

The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When $n=1$ ,

$(1+x)^1=1+x=\dbinom10+\dbinom11x$

so the base case holds. Assume the claim holds for $n=k$ , so that

$(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k$

Use this to show that it holds for $n=k+1$ .

$(1+x)^{k+1}=(1+x)(1+x)^k$
$(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)$
$(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}$

Notice that

$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}$

So you can write the expansion for $n=k+1$ as

$(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}$

and since $\dbinom{k+1}0=\dbinom{k+1}{k+1}=1$ , you have

$(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}$

and so the claim holds for $n=k+1$ , thus proving the claim overall that

$(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n$

Setting $x=1$ gives

$(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n$

which agrees with the result obtained for part (c).

4 0

3 years ago