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3 years ago

In ΔOPQ, m∠O = (3x+12)^{\circ}(3x+12)

Mathematics

1 answer:

zaharov [31]3 years ago

6 0

Answer:

The measure of angle P is 80 degrees

Step-by-step explanation:

we know that

The sum of the interior angles in a triangle must be equal to 180 degrees

In this problem we have that

$m\angle O+m\angle P+m\angle Q=180^o$

substitute the given values

$(3x+12)^o+(4x+4)^o+(x+12)^o=180^o$

solve for x

$(8x+28)=180$

$8x=180-28$

$8x=152$

$x=19$

<em>Find the measure of angle P</em>

$m\angle P=(4x+4)^o$

substitute the value of x

$m\angle P=(4(19)+4)=80^o$

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3 years ago

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Quadrilateral PAID is a rectangle whose diagonals have the endpoints P(-3, -2)I(4, -7) and A(4, -2)D(-3, -7). Find the diagonals

BARSIC [14]

Answer:

$(0.5,-4.5)$

Step-by-step explanation:

The given rectangle has diagonals have the endpoints P(-3, -2) ,I(4, -7) and A(4, -2) ,D(-3, -7)

The diagonals of the rectangle bisect each other so we use the midpoint formula to find their point of intersection.

The midpoint formula is;

$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

We use any pair of endpoints of the diagonals to find the point of intersection.

Using A(4, -2) ,D(-3, -7)

$(\frac{4+-3}{2},\frac{-2+-7}{2})$

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3 years ago

What percent is16 out of 64? It is _ .

Simora [160]

16 out of 64 is 25% hope this helped

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3 years ago

The n term of a geometric sequence is denoted by Tn and the sum of the first n terms is denoted by Sn.Given T6-T4=5/2 and S5-S3=

Leno4ka [110]

1 step: $S_{5}=T_{1}+T_{2}+T_{3}+T_{4}+T_{5}$ , $S_{3}=T_{1}+T_{2}+T_{3}$ , then
$S_{5}-S_{3}=T_{4}+T_{5}=5$ .

2 step: $T_{n}=T_{1}*q^{n-1}$ , then
$T_{6}=T_{1}*q^{5}$
$T_{5}=T_{1}*q^{4}$
$T_{4}=T_{1}*q^{3}$
$T_{3}=T_{1}*q^{2}$
and $\left \{ {{T_{6}-T_{4}= \frac{5}{2} } \atop {T_{5}+T_{4}=5}} \right.$ will have form $\left \{ {{T_1*q^{5}-T_{1}*q^{3}= \frac{5}{2} } \atop {T_{1}*q^{4}+T_{1}*q^{3}=5} \right.$ .

3 step: Solve this system $\left \{ {{T_1*q^{3}*(q^{2}-1)= \frac{5}{2} } \atop {T_{1}*q^{3}*(q+1)=5} \right.$ and dividing first equation on second we obtain $\frac{q^{2}-1}{q+1}= \frac{ \frac{5}{2} }{5}$ . So, $\frac{(q-1)(q+1)}{q+1} = \frac{1}{2}$ and $q-1= \frac{1}{2}$ , $q= \frac{3}{2}$ - the common ratio.

4 step: Insert $q= \frac{3}{2}$ into equation $T_{1}*q^{3}*(q+1)=5$ and obtain $T_{1}* \frac{27}{8}*( \frac{3}{2}+1 ) =5$ , from where $T_{1}= \frac{16}{27}$ .

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3 years ago

error uses 1/3 of wet dog food for his dog Muddy,each day. how many servings will he get from 5 cans of dog food.?

oksian1 [2.3K]

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3 years ago