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3 years ago

13

the volume of a shipping container is 50.148feet how many smaller boxes each with a volume of 2.786 cubic feet does the shipping

container hold

1 answer:

Bess [88]3 years ago

3 0

The answer is 18 I hope this helps

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I'll give brainliest

Tpy6a [65]

Given:

$\angle a$ and $\angle b$ are complementary angles.

$\angle a=32^\circ$

To find:

The measure of $\angle b$ .

Solution:

According to the definition of the complimentary angles, the sum of two complementary angles is always 90 degree.

It is given that, $\angle a$ and $\angle b$ are complementary angles. So,

$\angle a+\angle b=90^\circ$

$32^\circ+\angle b=90^\circ$

$\angle b=90^\circ-32^\circ$

$\angle b=58^\circ$

Therefore, the measure of $\angle b$ is 58°.

3 0

3 years ago

In Usama's high school, there are 190 teachers and 2650 students. What is the student-teacher ratio?

GarryVolchara [31]

Answer:

i think it's 14:1 i can't understand 503500.

6 0

3 years ago

Find all pairs of real numbers (a,b) such that (x-a)^2+(2x-b)^2=(x-3)^2+(2x)^2

Mrrafil [7]

Answer:

$(3,0)$

$(\frac{-9}{5},\frac{12}{5})$

Step-by-step explanation:

Let's expand both sides.

I'm going to use the following identity to expand the binomial squared expressions: $(u+v)^2=u^2+2uv+v^2$ or $(u-v)^2=u^2-2uv+v^2$ .

Left-hand side:

$(x-a)^2+(2x-b)^2$

$(x^2-2ax+a^2)+((2x)^2-2b(2x)+b^2)$

$x^2-2ax+a^2+4x^2-4bx+b^2$

Reorder so $x^2$ 's are together and that $x[tex]'s are together.[tex](x^2+4x^2)+(-2ax-4bx)+(a^2+b^2)$

$5x^2+(-2a-4b)x+(a^2+b^2)$

Right-hand side:

$(x-3)^2+(2x)^2$

$x^2-2(3)x+9+4x^2$

$x^2-6x+9+4x^2$

Reorder so $x^2$ 's are together and that $x[tex]'s are together.[tex](x^2+4x^2)+(-6x)+9$

$5x^2-6x+9$

Now let's compare both sides.

If we want both sides to appear exactly the same we need to choose values $a$ and $b$ such the following are true equations:

$-2a-4b=-6$

$a^2+b^2=9$

So if we solve the system we can find the values $a$ and $b$ such that the left=right.

Let's solve the first equation for $a$ in terms of $b$ .

Add $2a$ on both sides:

$-4b=-6+2a$

Divide both sides by -4:

$b=\frac{-6+2a}{-4}$

Reduce (divide top and bottom by -2):

$b=\frac{3-a}{2}$

Now let's plug this into second equation:

$a^2+b^2=9$

$a^2+(\frac{3-a}{2})^2=9$

$a^2+\frac{9-6a+a^2}{4}=9$ (I used the identity $(u-v)^2=u^2-2uv+v^2$ )

Multiply both sides by 4 to clear the fractions from the problem:

$4a^2+(9-6a+a^2)=36$

Combine like terms on left hand side:

$4a^2+a^2-6a+9=36$

$5a^2-6a+9=36$

Subtract 36 on both sides:

$5a^2-6a-27=0$

Now let's try to factor.

We are going to try to find two numbers that multiply to be 5(-27) and add to be -6.

5(-27)=(5*3)(-9)=15(-9)=-15(9) while -15+9=-6.

So let's replace $-6a$ with $-15a+9a$ and factor by grouping.

$5a^2-15a+9a-27=0$

$5a(a-3)+9(a-3)=0$

$(a-3)(5a+9)=0$

This implies $a-3=0$ or $5a+9=0$ .

Solving the first is easy. Just ad 3 on both sides to get: $a=3$ .

The second requires two steps. Subtract 9 and then divide by 5 on both sides.

$5a=-9$

$a=\frac{-9}{5}$ .

So let's go back to finding $b$ now that we know the $a$ values.

If $a=3$ and $b=\frac{3-a}{2}$ ,

then $b=\frac{3-3}{2}=0$ .

So one ordered pair $(a,b)$ that satisfies the equation is:

$(3,0)$ .

If $a=\frac{-9}{5}$ and $b=\frac{3-a}{2}$ ,

then $b=\frac{3-\frac{-9}{5}}{2}$ .

Let's multiply top and bottom by 5 to clear the mini-fraction.

$b=\frac{15-(-9)}{10}$

$b=\frac{24}{10}$

$b=\frac{12}{5}$

So one ordered pair $(a,b)$ that satisfies the equation is:

$(\frac{-9}{5},\frac{12}{5})$ .

5 0

4 years ago

put the number eight million, two hundred eleven thousand, four hundred nine into the place value chart, then write the number i

ipn [44]

Here are the answers

3 0

3 years ago

A function is given. Determine the average rate of change of the function between the given values of the variable: f(x)= 4x^2;

nevsk [136]

$\bf slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby 
\begin{array}{llll}
average\ rate\\
of\ change
\end{array}\\\\
-------------------------------$

$\bf f(x)= 4x^2 \qquad 
\begin{cases}
x_1=3\\
x_2=3+h
\end{cases}\implies \cfrac{f(3+h)-f(3)}{(3+h)~-~(3)}
\\\\\\
\cfrac{[4(3+h)^2]~~-~~[4(3)^2]}{\underline{3}+h-\underline{3}}\implies \cfrac{4(3^2+6h+h^2)~~-~~4(9)}{h}
\\\\\\
\cfrac{4(9+6h+h^2)~~-~~36}{h}\implies \cfrac{\underline{36}+24h+4h^2~~\underline{-~~36}}{h}
\\\\\\
\cfrac{24h+4h^2}{h}\implies \cfrac{\underline{h}(24+4h)}{\underline{h}}\implies 24+4h$

8 0

3 years ago