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2 years ago

Are the triangles similar? If so state the similarity and the postulate or theorem that justifies your answer.

Mathematics

1 answer:

tester [92]2 years ago

7 0

Answer:

Step-by-step explanation:

∆ABC is similar to ∆DEF if and only if the ratio of their corresponding side lengths are the same.

Let's find out:

DE/AB = 10/5 = 2

EF/BC = 8/4 = 2

FD/CA = 12/6 = 2

The ratio of their corresponding side lengths are equal, therefore, ∆ABC is similar to ∆DEF based on the SSS Similarity Theorem.

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What is the value of the expression 6z2+2z+10 when z=3?

grigory [225]

Hi there! Hope this helps, and if it does, please mark brainliest!

Answer:

346

Step-by-step explanation:

6z2+2z+10

(6 x 3)^2 + 2 x 3 + 10

18^2 + 12 + 10

324 + 22

346

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3 years ago

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Step-by-step explanation:

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2 years ago

Solve the following expression when : p=15. p/3+4

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2 years ago

Given that x + y = 13 and xy = 24, find the distance from the point (x, y) to the origin.

boyakko [2]

Answer: Distance from the point (x, y) to the origin is approximately 11 units.

Step-by-step explanation: Given equations x+y=13 and xy =24.

Solving first equation for y, we get

y = 13-x.

Substituting y=13-x in second equation, we get

x(13-x)= 24.

13x -x^2=24.

-x^2+13x =24.

-x^2+13x -24=0.

Dividing each term by -1, we get

x^2-13x+24=0.

Applying quadratic formula

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=1,\:b=-13,\:c=24:\quad x=\frac{-\left(-13\right)\pm \sqrt{\left(-13\right)^2-4\cdot \:1\cdot \:24}}{2\cdot \:1}$

$x=\frac{-\left(-13\right)+\sqrt{\left(-13\right)^2-4\cdot \:1\cdot \:24}}{2\cdot \:1}:\quad \frac{13+\sqrt{73}}{2} =10.77$

$x=\frac{-\left(-13\right)-\sqrt{\left(-13\right)^2-4\cdot \:1\cdot \:24}}{2\cdot \:1}:\quad \frac{13-\sqrt{73}}{2}=2.23$

Plugging x=10.77 in first equation

y= 13-10.77 = 2.23

and plugging x=2.23 in first equation, we get

y = 13-2.23 = 10.77.

Therefore, (x,y) are (10.77, 2.23) and (2.23, 10.77).

Now, we need to find the distance of (x,y) from origin (0,0).

Applying distance formula :

$\mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$=\sqrt{\left(10.77-0\right)^2+\left(2.23-0\right)^2}$

$=10.9984\$

<h3>≈ 11 units.</h3><h3>Therefore, distance from the point (x, y) to the origin is approximately 11 units.</h3>

3 0

3 years ago