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2 years ago

8

Fourteen friends went to the county fair. All except 6 of them bought a hot dog.each hot cost $3.how much did the friends spend

on hot dogs.

1 answer:

yanalaym [24]2 years ago

6 0

14 friends - 6 friends = 8
8 (friends) x 3 (dollars) = 24 (dollars total)
ANSWER = $24

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2) Given: lines a and b are parallel<br> Prove: m<1 = m<8

monitta

Explanation:

<u><em>* Demonstrate that the related angles are the same.</em></u>

<u><em>* Demonstrate that different interior angles are equivalent.</em></u>

<u><em>* Demonstrate the adjacent interior angles are supplementary.</em></u>

<u><em>* Demonstrate the adjacent external angles are not mutually exclusive.</em></u>

<u><em>* Show that the lines in a plane are perpendicular to the same axis.</em></u>

<u><em></em></u>

<1 and <8 are the same because they are alternate exterior angles that means that they are

Opposite sides of the transversal.

And that they are on the exterior side of the parallel lines.

Now because of that they have the same measures no matter what so would

<2 and <7

<3 and <6

<4 and <5

<u><em></em></u>

6 0

2 years ago

Two models of cellular telephones, red and blue, are stored in boxes. One box weighs twelve pounds and contains four of the red

GenaCL600 [577]

First, make a equation in which r= red and b=blue.

So, since in the first box it has 4 red models and one blue model equaling 12,

the first equation looks like 4r+b=12.

The second equation looks like 2r+b=8.

What you would do is try and first solve for r by getting rid of b.

Since both equation has a positive b, you would make one equation have a negative b by multiplying the whole equation by -1.

-1*(2r+b=8)= -2r-b=-8

Add.

4r+b=12
+(-2r-b=-8)

2r=4
r=2

Then, you plug in 2 for the r for any of the original equations.

4(2)+b=12

8+b=12

b=4

or

2(2)+b=8

4+b=8
b=4

So, the red models weigh 2 pounds while the blue models weigh 4.

5 0

3 years ago

Multiply the following rational expressions and simplify the result

GarryVolchara [31]

Answer:

Step-by-step explanation:

We have to solve the given expression,

$\frac{9y-33y^2-3y^4}{100-49y^2}.\frac{7y^2+17y+10}{14y^2+28y}$

$\frac{9y-33y^2-3y^4}{100-49y^2}.\frac{7y^2+17y+10}{14y^2+28y}$ = $\frac{-y(-9+33y+3y^3)}{100-49y^2}.\frac{7y^2+17y+10}{14y(y+2)}$

= $\frac{-y(-9+33y+3y^3)}{(10-7y)(10+7y)}.\frac{7y^2+10y+7y+10}{14y(y+2)}$

= $\frac{-y(-9+33y+3y^3)}{(10-7y)(10+7y)}.\frac{y(7y+10)+1(7y+10)}{14y(y+2)}$

= $\frac{-y(-9+33y+3y^3)}{(10-7y)(10+7y)}.\frac{(y+1)(7y+10)}{14y(y+2)}$

= $\frac{-3y(-3+11y+y^3)}{(10-7y)}.\frac{(y+1)}{14y(y+2)}$

= $\frac{-3(-3+11y+y^3)}{(10-7y)}.\frac{(y+1)}{14(y+2)}$

= $\frac{3(3-11y-y^3)(y+1)}{(10-7y)(14(y+2)}$

3 0

3 years ago

Match the circle equations in general form with their corresponding equations in standard form. Not all will be used.

Xelga [282]

<span>The standard form of the equation of a circumference is given by the following expression:

</span> $(x-h)^{2}+(y-k)^{2}=r^{2} \\ \\ where \ (h, k) \ is \ the \ center \ of \ the \ circumference \ and \ r \ the \ radius$
<span>
On the other hand, the general form is given as follows:

</span> $x^{2}+y^{2}+Dx+Ey+F=0 \\ \\ where: \\ D=-2h, \ E=-2k, \ F=h^{2}+k^{2}-r^{2}$ <span>

In this way, we can order the mentioned equations as follows:

Equations in Standard Form:

</span> $\bold{a)} \ (x-6)^{2}+(y-4)^{2}=56 \\ \bold{b)} \ (x-2)^{2} + (y+6)^{2}=60 \\ \bold{c)} \ (x+2)^{2}+(y+3)^{2}=18 \\ \bold{d)} \ (x+1)^{2}+(y-6)^{2}=46$

Equations in General Form:

$\bold{1)} \ x^{2}+y^{2}-4x+12y-20=0 \\ \bold{2)} \ x^{2}+y^{2}+6x-8y-10=0 \\ \bold{3)} \ 3x^{2}+3y^{2}+12x+18y-15=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 3, \ the \ equation \ becomes: \\ x^{2}+y^{2}+4x+6y-5=0 \\ \\ \bold{4)} \ 5x^{2}+5y^{2}-10x+20y-30=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 5, \ the \ equation \ becomes: \\ x^{2}+y^{2}-2x+4y-6=0 \\ \\ \bold{5)} \ 2x^{2}+2y^{2}-24x-16y-8=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 2, \ the \ equation \ becomes: \\ x^{2}+y^{2}-12x-8y-4=0$

$\bold{6)} \ x^{2}+y^{2}+2x-12y$

So let's match each equation:

$\bold{From \ a)} \\ \\ (h,k)=(6,4),\ r=2\sqrt{14} \\ D=-12, \ E=-8 \\ F=-4$

Then, its general form is:

$x^{2}+y^{2}-12x-8y-4=0$

<em><u>First. a) matches 5) </u></em>

$\bold{From \ b)} \\ \\ (h,k)=(2,-6),\ r=2\sqrt{15} \\ D=-4, \ E=12 \\ F=-20$

Then, its general form is:

$x^{2}+y^{2}-4x+12y-20=0$

<em><u>Second. b) matches 1) </u></em>

$\bold{From \ c)} \\ \\ (h,k)=(-2,-3),\ r=3\sqrt{2} \\ D=4, \ E=6 \\ F=-5$

Then, its general form is:

$x^{2}+y^{2}+4x+6y-5=0$

<em><u>Third. c) matches 3)</u></em>

$\bold{From \ d)} \\ \\ (h,k)=(-1,6),\ r=\sqrt{46} \\ D=2, \ E=-12, \ F=-9$

Then, its general form is: $x^{2}+y^{2}+2x-12y-9=0$

<em><u>Fourth. d) matches 6)</u></em>

6 0

3 years ago

Read 2 more answers

The product of 5 and a number less 6 is 4. write the phrase as an algebraic equation.

kvasek [131]

5 + x - 6 = 4
x= 5

5 plus 5 equals 10 and 10 minus 6 is equal to 4.

3 0

2 years ago

Read 2 more answers