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2 years ago

The figure to the right has its congruent segments marked. Find the measure of AB.

Mathematics

1 answer:

Vaselesa [24]2 years ago

5 0

Answer:

the measure of AB =29 meters

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Find the solution of the system of equations.<br><br> -x+2y=-15<br> 8x-2y=-20

Grace [21]

Answer:

the answer is (-5,-10)

hope this helps :)

8 0

2 years ago

Which is the more reasonable measurement of the distance between the tracks on a railroad: 1.435 ⋅ 10−3 mm or 1.435 ⋅ 103 mm? Ju

SVEN [57.7K]

The best answer is 1.435 ·103 because while they both are really small for a rail road track that one is bigger. the final answer for that is 147.805mm.

5 0

2 years ago

Read 2 more answers

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

2 years ago

Sandra knows the Pythagorean identity sin 2 ⁡ θ + cos 2 ⁡ θ = 1 . If she is told that 0 ≤ θ ≤ π 2 and cos ⁡ θ = 5 12, what will

slavikrds [6]

Answer:

$\text{tan}(\theta)=\frac{\sqrt{119}}{5}$

Step-by-step explanation:

We have been given that Sandra knows the Pythagorean identity $\text{sin}^2(\theta)+\text{cos}^2(\theta)=1$ . She is told that $0\leq \theta\leq \frac{\pi}{2}$ and $\text{cos}(\theta)=\frac{5}{12}$ .