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3 years ago

A triangle has angle measurements of 30°, 70°, and 80°. What kind of triangle is it?

Mathematics

1 answer:

Juliette [100K]3 years ago

5 0

The answer is acute triangle because none of the angles are greater than 90°

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Which answer would be correct?<br><br> is it 7.23 or something else?<br><br> * ANSWER ASAP !!! *

vladimir2022 [97]

<h3>Answer: Choice B) 8.57</h3>

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Explanation:

The triangles are similar through the AA (angle angle) similarity rule. Note how the angle markers correspond and match up. Example: angle C and angle F both have triple angle markers to indicate these angles are congruent.

The corresponding sides pair up to form equal fractions. AB and DE form AB/DE which is equal to AC/DF, as these two sides correspond as well.

Therefore: AB/DE = AC/DF

Let's plug in the given values and isolate x

AB/DE = AC/DF

11/7 = (15+x)/15 ...... substitution

11*15 = 7*(15+x) ....... cross multiply

11*15 = 7*15+7*x ...... distribute

165 = 105+7x

165-105 = 105+7x-105 ..... subtract 105 from both sides

60 = 7x

7x = 60

7x/7 = 60/7 ............ divide both sides by 7

x = 8.57142857142858

x = 8.57 ..... rounding to the nearest hundredth

3 0

3 years ago

Read 2 more answers

What is the surface area of a sphere with radius 2?

VladimirAG [237]

$\textit{surface area of a sphere}\\\\ SA=4\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=2 \end{cases}\implies \begin{array}{llll} SA=4\pi (2)^2\implies SA=16\pi \\\\\\ SA\approx 50.27 \end{array}$

8 0

2 years ago

Solve the inequality 2x>30+5/4x

insens350 [35]

Answer:

Step-by-step explanation:

$2x > 30+\frac{5}{4x} \\2x-\frac{5}{4x} > 30\\\frac{8x^2-5}{4x} > 30\\case~1\\if~x > 0\\8x^2-5 > 120x\\8x^2-120x > 5\\x^2-15x > \frac{5}{8} \\adding~(-\frac{15}{2} )^2~to~both~sides\\(x-\frac{15}{2} )^2 > \frac{5}{8}+\frac{225}{4} \\(x-\frac{15}{2} )^2 > \frac{455}{8} \\x-\frac{15}{2} < -\sqrt{\frac{455}{8} } \\x < \frac{15}{2}-\sqrt{\frac{455}{8} } \\or~x < 0\\rejected~as~x > 0$

$x-\frac{15}{2} > \sqrt{\frac{455}{8} } \\x > \frac{15}{2} +\sqrt{\frac{455}{8} }$

case~2

$if~x < 0\\8x^2-5 < 120x\\8x^2-120x < 5\\x^2-15x < \frac{5}{8} \\adding~(-\frac{15}{2} )^2\\(x-\frac{15}{2} )^2 < \frac{5}{8} +(-\frac{15}{2} )^2\\|x-\frac{15}{2} | < \frac{5+450}{8} \\-\sqrt{\frac{455}{8} } < x-\frac{15}{2} < \sqrt{\frac{455}{8} } \\\frac{15}{2} -\sqrt{\frac{455}{8} } < x < \frac{15}{2} +\sqrt{\frac{455}{8} } \\but~x < 0\\7.5-\sqrt{\frac{455}{8} } < x < 0$

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2 years ago

Find the value of x and y that maximize or minimize the objective function.

kogti [31]

0 > x > 2 and y < 8 - x

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3 years ago

I beeps help please

Sergeeva-Olga [200]

Answer is in the file

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3 years ago