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2 years ago

THE QUESTION IS EASY PLS ANSWER THE QUESTION . BUT I DONT KNOW PLSS

Mathematics

2 answers:

Sliva [168]2 years ago

5 0

Well you can determine the unknown angle in the triangle by knowing that a triangles angles add up to 180.
64 + 52 + (Unknown) = 180
Unknown = 64
Unknown and the angle of F create a 180 angle relationship. Therefore,
F = 180-64 —> 116
Angle D has a 180 angle relationship with F and so does X with D, so F and X are the same.
Answer: 116

Shkiper50 [21]2 years ago

3 0

Answer:

x=116

Step-by-step explanation:

180-C will get you F since they are adjacent angles, which would be 116. We know that the opposite angles of a parallelogram are equal, so x=116

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Find the value of y when x equals 0 . -7x+3y=30

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3 years ago

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Solve the equation 2.7 + x = 17

olga55 [171]

Hello!

To solve this equation, we need to isolate the variable on one side of the equation to see what it equals.

As you can see, 2.7 is added. To get it onto the other side and cancel it from the one side, you will subtract it from both sides as shown below. Note that because of the commutative property, we can switch the places of 2.7 and x.

x+2.7-2.7=17-2.7 (Note that 2.7=2.7 is 0 in this equation.
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2 years ago

Which beat describes how to solve the equation below? 35x=88 A. Divide both sides by 88 B. Multiply both sides by 35 C. Divide b

Alex Ar [27]

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2 years ago

Given csc(A) = 65/16 and that angle A is in Quadrant I, find the exact value of sec A in simplest radical form using a rational

charle [14.2K]

Answer:

$secA = \frac{65}{63}$

Step-by-step explanation:

$cosec A = \frac{65}{16}\\\\sin A = \frac{1}{cosecA} = \frac{16}{65}\\\\cos^2 A = 1 - sin^2 A$

$= 1 - (\frac{16}{65})^2\\\\=\frac{4225-256}{4225}\\\\=\frac{3969}{4225}\\$

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2 years ago

Triangle pqr has vertices p(-3 -1) q(-1,-7) and r(3,3) and points a and b are midpoints of segment pq and segment rq respectivel

Serhud [2]

Answer:

View graph

Step-by-step explanation:

Be: p(-3,-1) q(-1,-7) r(3,3)

we must find the value of the middle point a and b

$a=\frac{pq}{2}\\b=\frac{rq}{2}$

remember that distance between two points is:

$d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} }$

pq=distance between p and q

qr =distance between q and r $pq=\sqrt{(-1+3)^{2}+(-7+1)^{2}}=\sqrt{(2)^{2}+(-6)^{2}}=\sqrt{40}\\ pq=2\sqrt{10}\\ a=\frac{pq}{2}=\frac{2\sqrt{10}}{2}=\sqrt{10}\\ coordinates\\a=\frac{pq}{2}=\frac{(-3-1,-7-1)}{2}=(-2,-4)\\\\qr=\sqrt{(3+1)^{2}+(3+7)^{2}}=\sqrt{(4)^{2}+(10)^{2}}=\sqrt{116}\\qr=2\sqrt{29}\\ b=\frac{qr}{2}=\frac{2\sqrt{29}}{2}=\sqrt{29}\\ coordinates\\b=\frac{pq}{2}=\frac{(-1+3,-7+3)}{2}=(1,-2)\\$

ab is parallel to pr if slope ab is equal to slope pr

$m_{ab}=\frac{-2+4}{1+2}=\frac{2}{3}\\m_{pr}=\frac{3+1}{3+3}=\frac{2}{3}\\ m_{ab}=m_{pr}$

finally distance ab is equal to distance pr/2

$d_{ab=}\sqrt{(-2+4)^{2}+(1+2)^{2}}=\sqrt{13}\\ d_{pr=}\sqrt{(3+1)^{2}+(3+3)^{2}}=\sqrt{52}\\ d_{ab=}\frac{d_{pr}}{2}$

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