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3 years ago

Please help me with this math problem

Mathematics

1 answer:

ollegr [7]3 years ago

8 0

$10 \times 5 + \frac{ {4}^{2} }{4} = 50 + 4 = 54$

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I F 49x ²b = (7x +1/2)(7x-1/2)<br>then Find the value b.

Marina86 [1]

Answer:

Step-by-step explanation:

Apply Only the Outside and Inside Method of the Foil Method.

$7 \times - \frac{1}{2} = - 3.5x$

$\frac{1}{2} \times 7x = 3.5x$

Add them together

$- 3.5x + 3.5x = 0$

So our b value is 0.

4 0

3 years ago

Help?pleaseeee?someone?anyone?

slamgirl [31]

Answer:

the answer is 96

Step-by-step explanation:

$(-10)^{2}$ - 9 $(\frac{1}{3} )^{2}$ -3

100- $\frac{9}{9}$ -3

100-1-3

6 0

3 years ago

Please and thank you

Alina [70]

The answer is -1 and 0

6 0

3 years ago

Read 2 more answers

Find the interquartile range for the box plot below.*

alexgriva [62]

Answer:

Step 1: Put the numbers in order. ...

Step 2: Find the median. ...

Step 3: Place parentheses around the numbers above and below the median. Not necessary statistically, but it makes Q1 and Q3 easier to spot. ...

Step 4: Find Q1 and Q3. ...

Step 5: Subtract Q1 from Q3 to find the interquartile range.

Step-by-step explanation:

u should do those step and i think u will find the answer

6 0

3 years ago

Please help solve this system of equations

stepan [7]

Make a substitution:

$\begin{cases}u=2x+y\\v=2x-y\end{cases}$

Then the system becomes

$\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}$

Simplifying the equations gives

$\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}$

which is to say,

$\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}$

$\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81$

$\implies\dfrac uv=\pm27$

$\implies u=\pm27v$

Substituting this into the new system gives

$\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1$

$\implies u=\pm27$

Then

$\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13$

(meaning two solutions are (7, 13) and (-7, -13))

8 0

3 years ago