Help please please please

Someone please help me this is the last question on my homework. I have to write a quadratic equation for this word problem...

antiseptic1488 [7]

Answer:

57xsqaure-42x=5.5

Step-by-step explanation:

5 0

3 years ago

Over the last 3 evenings, Keisha received a total of 81 phone calls at the call center. The first evening, she received 6 few

tensa zangetsu [6.8K]

Answer:

First: 15

Second: 21

Third: 45

Step-by-step explanation:

In order to create the equation we need to represent the evenings with an alebraic term, in thsi case we are going to represent the second evening with an X

Second evening:x

The first night she got 6 fewer calls than the second: Second-6=x-6

The third night she received 3 times the first: 3(first night)=3(x-6)

The equation is First Night plus second night plus third night equals 81.

$First+second+third=81\\x+(x-7)+3(x-6)=81\\x+x-7+3x-18=81\\5x=81+21+6\\5x=105\\x=\frac{105}{5} \\x=21$

So the first evening he received 15 calls, the second he received 21 and the third one he received 45.

8 0

3 years ago

Help ASAP!!!! Thanks

Tema [17]

Answer:

No DB is not a perpendicular bisector of AC

Step-by-step explanation:

This is because as AC is a straight line it's angle degree is 180 which when bisected by DB becomes,

180 ÷ 2 = 90

On both the angles i.e <BDC = <NDA = 90°

To make it a perpendicular bisector but

<BDC is not equal to <NDA is not equal to 90°.

Hence, DB is not a perpendicular bisector of line AC.

8 0

3 years ago

What is the value of x? x = ?

meriva

 Sum of both the angle will 180 because of linear pair

now, 10x - 20 + 6x + 8 =. 180 

 16x - 12 = 180

16x = 192 

x= 192/16

x = 12

1st angle = 100

2nd is = 80

Hope it helps and your day will full of happiness

8 0

3 years ago

Read 2 more answers

The perimeter of a triangular field is 120m. Two of the sides are 21m and 40m.Calculate the largest angle of the field.

Mama L [17]

Answer:

the largest angle of the field is 149⁰

Step-by-step explanation:

Given;

perimeter of the triangular filed, P = 120 m

length of two known sides, a and b = 21 m and 40 m respectively

The length of the third side is calculated as follows;

a + b + c = P

21 m + 40 m + c = 120 m

61 m + c = 120 m

c = 120 m - 61 m

c = 59 m

B

↓ ↓

A → → → → → → → → → → → C

Consider ABC as the triangular field;

Angle A is calculated by applying cosine rule;

$a^2 = b^2 + c^2 - 2bc \ Cos A\\\\Cos \ A = \frac{b^2 + c^2 - a^2}{2bc} \\\\Cos \ A = \frac{40^2 + 59^2 - 21^2}{2 \times 40 \times 59} \\\\Cos \ A = 0.983\\\\A = Cos ^{-1} (0.983)\\\\A = 10.6 \ ^0$

Angle B is calculated as follows;

$Cos \ B = \frac{a^2 + c^2 - b^2}{2ac} \\\\Cos \ B = \frac{21^2 + 59^2 - 40^2}{2 \times 21 \times 59} \\\\Cos \ B = 0.937\\\\B= Cos ^{-1} (0.937)\\\\B = 20.5 \ ^0$

Angle C is calculated as follows;

$Cos \ C = \frac{a^2 + b^2 - c^2}{2ab} \\\\Cos \ C = \frac{21^2 + 40^2 - 59^2}{2 \times 21 \times 40} \\\\Cos \ C = -0.857\\\\C = Cos ^{-1} (-0.857)\\\\C = 149\ ^0$

Therefore, the largest angle of the field is 149⁰.

8 0

3 years ago