do you have any more information?
Answer:
PT and SQ
Step-by-step explanation:
IF two segments are perpendicular then they will create 90 degree angles
segment PR and segment SQ cross over each other and create 90 degree angles therefore segment PR is perpendicular to segment SQ
<span>This is the equation made from the problem where x=mystery number
</span><span>2x+3(x+1)=4(x-1)</span><span>
</span><span>Now let's solve for x!
</span><span>
</span><span>We start by distributing 3 into (X+1)
</span><span>
</span><span>3(x)=3x and 3(1)=3
</span><span>
</span><span>Now our equation is 2x+3x+3=4(x-1)
</span><span>
</span><span>Let's combine both x values on the left side of the equation: 2x + 3x=5x
</span><span>
</span><span>We now have 5x+3=4(x-1)
</span><span>
</span><span>Let's distribute 4 into (x-1)
</span><span>
</span><span>4(x)=4x and 4(-1)=-4
</span><span>
</span><span>Now our equation is 5x+3=4x-4
</span><span>
</span><span>subtract 3 form both sides
</span><span>
</span><span>5x=4x-7
</span><span>
</span><span>subtract 4x from both sides
</span><span>
</span><span>x=-7
</span><span>
</span><span>Yay! So the number she is thinking of is -7!</span><span>
</span>
Answer:
<h2><em>
38°, 66° and 76°</em></h2>
Step-by-step explanation:
A triangle consists of 3 angles and sides. The sum of the angles in a triangle is 180°. Let the angle be <A, <B and <C.
<A + <B + <C = 180° ...... 1
If the measure of one angle is twice the measure of a second angle then
<A = 2<B ...... 2
Also if the third angle measures 3 times the second angle decreased by 48, this is expressed as <C = 3<B-48............ 3
Substituting equations 2 and 3 into 1 will give;
(2<B) + <B + (3<B-48) = 180°
6<B- 48 = 180°
add 48 to both sides
6<B-48+48 = 180+48
6<B = 228
<B = 228/6
<B =38°
To get the other angles of the triangle;
Since <A = 2<B from equation 2;
<A = 2(38)
<A = 76°
Also <C = 3<B-48 from equation 3;
<C = 3(38)-48
<C = 114-48
<C = 66°
<em>Hence the measures of the angles of the triangle are 38°, 66° and 76°</em>
What about that??? I don't what to solve for