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3 years ago

12

Solve for y on problem #2

1 answer:

Anestetic [448]3 years ago

5 0

9514 1404 393

Answer:

y = (x +7)/3

Step-by-step explanation:

Multiply the given equation by y/3:

$\dfrac{x+7}{y}\cdot\dfrac{y}{3}=3\cdot\dfrac{y}{3}\\\\\boxed{y=\dfrac{x+7}{3}}$

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8^10 x 8^4 simplified

Mandarinka [93]

Answer:

8^14

Step-by-step explanation:

8^10 * 8^4

We know that a^b * a^c = a^(b+c)

8^(10+4)

8^14

6 0

3 years ago

Read 2 more answers

The town of yarmouth is planning a skateboard park and needs to know the perimeter of the park. The property measures 7 yards by

fiasKO [112]

To get the perimeter, you get the sum of all the sides.
Therefore, the perimeter of the park here is 7+3+10+5 = 25 yards.

8 0

3 years ago

The graph of linear function f passes through the point (1 -9) and has a slope of -3.

AleksandrR [38]

So think of it as division divide 3 by 2 and your anwser is what you hav

8 0

3 years ago

RHOMBUS The diagonals of rhombus ABCD intersect at E. Given that

schepotkina [342]

Answer:

<u>Question 11:</u>

$\angle DAC = 53^\circ$

$\angle AED = 90^\circ$

$\angle ADC = 74$

$DB = 16$

$AE = 6.03$

$AC = 12.06$

<u>Question 12:</u>

$\triangle ABD$ , $\triangle BAC$ , $\triangle CDA$ and $\triangle DAB$

<u>Question 13: </u>

AC and BD are perpendicular lines, and they are diagonals

Step-by-step explanation:

<u>Question 11</u>

Given

$\angle BAC = 53^\circ$

$DE = 8$

See attachment for Rhombus

Required

Determine the indicated sides

Solving (a): $\angle DAC$

Diagonal CA divides $\angle DAB$ into 2 equal angles

i.e

$\angle DAC = \angle BAC$

So:

$\angle DAC = 53^\circ$

Solving (b): $\angle AED$

The angles at E is 90 degrees because diagonals AC and BD meet at a perpendicular.

So:

$\angle AED = 90^\circ$

Solving (c): $\angle ADC$

First, we calculate $\angle ADE$ , considering $\triangle ADE$ :

$\angle ADE + \angle AED + \angle DAC = 180$

$\angle ADE + 90 + 53 = 180$

$\angle ADE + 143 = 180$

$\angle ADE = -143 + 180$

$\angle ADE = 37$

To calculate $\angle ADC$ , we have:

$\angle ADC = 2*\angle ADE$

$\angle ADC = 2* 37$

$\angle ADC = 74$

Solving (d): $DB$

From the rhombus

$DB = DE +EB$

Where

$DE =EB$

So:

$DB = 8 + 8$

$DB = 16$

Solving (e): $AE$

To do this we consider $\triangle ADE$

Using the tan formula

$tan(\angle ADE) = \frac{AE}{DE}$

$\angle ADE = 37$ and $DE = 8$

So:

$\tan(37) = \frac{AE}{8}$

$AE = 8 * \tan(37)$

$AE = 6.03$

Solving (f): $AC$

This is calculated as:

$AC = AE + EC$

Where

$AE = EC$

$AC = 6.03 +6.03$

$AC = 12.06$

<u>Question 12: Isosceles Triangle</u>

In the rhombus, all 4 sides are equal;

So, the isosceles triangle are:

$\triangle ABD$ , $\triangle BAC$ , $\triangle CDA$ and $\triangle DAB$

<u>Question 13: </u>

AC and BD are perpendicular lines, and they are diagonals

4 0

3 years ago

Please this IXL is due tomorrow I really am struggling.

balandron [24]

Number of people can hold in each car = 10

Solution:

Number of cars in the Blazing Bullet = 6

Total number of people can ride at one time = 60

Number of people can hold at one Bullet car = ?

Let n be the number of people in each car can hold.

Now write a equation to find the number of people in each car.

Number of cars × number of people in each car = Total number of people

⇒ 6 × n = 60

⇒ 6n = 60

Now, solving this equation.

⇒ 6n = 60

Divide both sides of the equation by 6.

$$ \Rightarrow\frac{6n}{6} = \frac{60}{6}$

$$ \Rightarrow n=10$

⇒ n = 10

Hence 10 people can hold in each car.

6 0

4 years ago