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3 years ago

Who knows this, Find m∠UTS, if m∠UTV=x+15, m∠VTS=140∘, and m∠UTS=15x+15

Mathematics

1 answer:

kakasveta [241]3 years ago

5 0

Answer:

165

Step-by-step explanation:

First, m<UTS=m<UTV+m<VTS by Angle Addition Postulate. Then, you substitute all the values that you provided for the angles. 15x+15=x+15+140. You then solve for x.

15x+15=x+155

14x=140

x=10

You then plug back in 10 for X in the value of m<UTS. 15(10)+15=165

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3 years ago

Determine the measure of each segment then indicate whether the statements are true or false

kupik [55]

Answer:

$d_{AB}\ne d_{JK}$

$d_{AB}\ne \:d_{GH}$

$d_{GH}\ne \:d_{JK}$

Therefore,

Option (A) is false

Option (B) is false

Option (C) is false

Step-by-step explanation:

Considering the graph

Given the vertices of the segment AB

A(-4, 4)

B(2, 5)

Finding the length of AB using the formula

$d_{AB}\:=\:\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$=\sqrt{\left(2-\left(-4\right)\right)^2+\left(5-4\right)^2}$

$=\sqrt{\left(2+4\right)^2+\left(5-4\right)^2}$

$=\sqrt{6^2+1}$

$=\sqrt{36+1}$

$=\sqrt{37}$

$d_{AB}\:=\sqrt{37}$

$d_{AB}=6.08$ units

Given the vertices of the segment JK

J(2, 2)

K(7, 2)

From the graph, it is clear that the length of JK = 5 units

$d_{JK}=5$ units

Given the vertices of the segment GH

G(-5, -2)

H(-2, -2)

Finding the length of GH using the formula

$d_{GH}\:=\:\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$=\sqrt{\left(-2-\left(-5\right)\right)^2+\left(-2-\left(-2\right)\right)^2}$

$=\sqrt{\left(5-2\right)^2+\left(2-2\right)^2}$

$=\sqrt{3^2+0}$

$=\sqrt{3^2}$

$\mathrm{Apply\:radical\:rule\:}\sqrt[n]{a^n}=a,\:\quad \mathrm{\:assuming\:}a\ge 0$

$d_{GH}\:=\:3$ units

Thus, from the calculations, it is clear that:

$d_{AB}=6.08$

$d_{JK}=5$

$d_{GH}\:=\:3$

Thus,

$d_{AB}\ne d_{JK}$

$d_{AB}\ne \:d_{GH}$

$d_{GH}\ne \:d_{JK}$

Therefore,

Option (A) is false

Option (B) is false

Option (C) is false

8 0

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2 years ago

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x+2-----> third positive integer

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using a graph tool-------> <span>I solve the quadratic equation
</span>see the attached figure

the roots are
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