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3 years ago

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A family of 2 adults and 3 children goes to a play. Admission cost $8 per adult and $5 per child. What expression would show the

total admission cost fir the family?

1 answer:

Reika [66]3 years ago

7 0

The answere to your question is $31 because 2 times 8 is 16 and 3 times 5 is 15 so if 15 plus 15 is 30 but sense it is 16 instead of 15 it is $31.

You might be interested in

Solving simultaneous equations with one quadratic, x^2+y^2=16 and y=x+3

Svetlanka [38]

If $y=x+3$ , then in the first equation you have

$x^2+(x+3)^2=x^2+(x^2+6x+9)=2x^2+6x+9=16$

Rewriting a bit, you get

$x^2+3x+\dfrac92=8$

$x^2+3x+\dfrac94+\dfrac94=8$

$\left(x+\dfrac32\right)^2=\dfrac{23}4$

$x=-\dfrac32\pm\dfrac{\sqrt{23}}2$

Now, since $y=x+3$ , you also get

$y=-\dfrac32\pm\dfrac{\sqrt{23}}2+3=\dfrac32\pm\dfrac{\sqrt{23}}2$

So, there are two solutions here:

$(x,y)=\left(-\dfrac32+\dfrac{\sqrt{23}}2,\dfrac32+\dfrac{\sqrt{23}}2\right)$
$(x,y)=\left(-\dfrac32-\dfrac{\sqrt{23}}2,\dfrac32-\dfrac{\sqrt{23}}2\right)$

7 0

3 years ago

the length of a rectangle is 50 feet less than its width. If the perimeter of the field is 840 feet, find the length of the fiel

LiRa [457]

$look\ at\ the\ picture\\\\2a+2(a-50)=840\\\\2a+2a-100=840\\\\4a-100=840\ \ \ /+100\\\\4a=940\ \ \ \ /:4\\\\a=235\ (ft)\\\\a-50=235-50=185\ (ft)\\\\Answer:lenght=185\ ft;\ width=235\ ft.$

8 0

3 years ago

E is the midpoint of Df, DE=2x+4, and EF=3x-1. Find all segment lengths

sertanlavr [38]

If E is the midpoint, DE and EF are equivalent.
2x+4=3x-1
4=x-1
5=x

Plug the value of x in
2(5)+4
14

3(5)-1
14

14+14=28

Final answer: DE=14, EF=14, DF= 28

6 0

4 years ago

Two cargo ships are sailing a direct course into harbor. Ship A is 22 miles from harbor. Ship Bis sailing into harbor on a cours

s344n2d4d5 [400]

Answer:

The distance of ship B from the Harbor is 32.26 miles

Step-by-step explanation:

Hello, great question. These types are questions are the beginning steps for learning more advanced Equations.

I have added a visual aid of the problem. As we can see the courses of both ship to the harbor forms a triangle with ship B being the hypotenuse. Since we are given the distance of the course for ship A and the angle between both ships, we can use this information with the <u>COSINE operator</u> to solve for the length of the course of ship B.

$cos(x) = \frac{adjacent}{hypotenuse}$

$cos(47) = \frac{22miles}{hypotenuse}$ .... flip both fractions.

$\frac{1}{cos(47)} = \frac{hypotenuse}{22miles}$ .... multiply both sides by 22

$\frac{22mi}{cos(47)} = hypotenuse$

$32.26mi = hypotenuse = B$

So the distance of ship B from the Harbor is 32.26 miles

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

3 0

3 years ago

Can someone help me understand this?

Irina18 [472]

Well if we have two angles, LMT and TMN, that make up LMN, then the sum of LMT and TMN should equal LMN.

144 + 23 = 167 degrees.

Hope this helps!

3 0

2 years ago