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3 years ago

A triangle has vertices at the points A(1,-5), B(-4,-3), and C(3,-3). What is the area of this triangle?

Mathematics

1 answer:

Makovka662 [10]3 years ago

7 0

Answer:

the answer to the question will be A

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What is the distance between -2.2 , 8.4 on a number line

leonid [27]

Answer:

10.2

Step-by-step explanation:

if u write is down with the decimals tgat and count that is what u would get also 2 plus 8 is ten and two pluse two is 41

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2 years ago

Plzzzzzzzzzzz answerrrrrrrrrrrrrrrr i need it asap

saul85 [17]

Answer: 120

1.2

If Maxine is correct, then she spent

1.2

times the hours she did homework than last week.

⋅

1.2

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This means that Maxine is correct.

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6 0

3 years ago

I need to know if it’s A B C OR D

Liula [17]

SOLUTION

Consider the property, opposite angle of a cyclic quadrilateral are supplementary

hence

$\angle NMP+\angle NOP=180^0$

Recall that

$\begin{gathered} \angle NMP=\angle M=(8x-24)^0 \\ \text{and } \\ \angle NOP=\angle O=(4x)^0 \end{gathered}$

Hnece, we have that

$\begin{gathered} \angle M+\angle O=180^0(opposite\text{ angles of a cyclic quadrilateral)} \\ (8x-24)^0+4x^0=180^0 \end{gathered}$

Then

$\begin{gathered} 8x-24+4x=180 \\ 8x+4x-24=180 \\ 12x-24=180 \\ \text{Add 24 t o both sides } \\ 12x-24+24=180+24 \\ 12x=204 \\ \text{divide both sides by 12} \\ \frac{12x}{12}=\frac{204}{12} \\ \text{Then} \\ x=17 \end{gathered}$

Hence

x=17

Since

$\begin{gathered} \angle NOP=(4x)^0 \\ \text{substitute the value of x, we have } \\ \angle NOP=4\times17=68^0 \\ \text{Hence } \\ \angle NOP=68^0 \end{gathered}$

Therefore

The measure of angle NOP is 68⁰

Answer; 68⁰ (The fourth Option )

5 0

1 year ago

I need help w/ this!! thank you

Fittoniya [83]

<h3>Answer:</h3>

$\boxed{\pink{\sf \leadsto Yes \ there \ is \ a \ solution \ of \ the \ given \ inequality .}}$

<h3>Step-by-step explanation:</h3>

A inequality is given to us and we need to convert it into standard form and see whether if it has a solution . So let's solve the inequality.

The inequality given to us is :-

$\bf\implies |2y + 3 | - 1 \leq 0 \\\\\bf\implies |2y+3|\leq 1 \\\\\bf\implies (|2y+3|)^2 \leq 1^2 \\\\\bf\implies (2y+3)^2 \leq 1 \\\\\bf\implies (2y)^2+3^2+2(2y)(3) \leq 1 \\\\\bf\implies 4y^2+9+12y - 1 \leq 0 \\\\\bf\implies 4y^2+12y+8 \leq 0 \\\\\bf\implies 4( y^2 + 3y + 2 ) \leq 0 \\\\\bf\implies y^2+3y +2 \leq 0 \:\:\bigg\lgroup \purple{\bf Standard \ form \ of \ inequality }\bigg\rgroup \\\\\bf\implies y^2y+2y+y+2 \leq 0 \\\\\bf\implies y(y+2)+1(y+2)\leq 0 \\\\\bf\implies ( y+2)(y+1)\leq 0 \\\\\bf\implies \boxed{\red{\bf y \leq (-2) , (-1) }}$