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lilavasa [31]
3 years ago
10

Can someone help me 10 points

Mathematics
1 answer:
Elanso [62]3 years ago
3 0

The two angles form a right triangle which is 90 degrees.

Add the two to equal 90:

3x-8 + 4x = 90

simplify:

7x-8 = 90

Add 8 to both sides:

7x = 98

Divide both sides by 7:

x = 14

You might be interested in
Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be
Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of t=0.6482

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

h(t)=-16t^2 +v_{o}t+s       Eqn(1).

Where:

h(t) : is the height range as a function of time

v_{o}   : is the initial velocity

s     : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

h(t)=-16t^2 + v_{o}t + 40        Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

<u>Ben:</u>

We know that v_{o}=60ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+60t+40        Eqn.(3)

<u>Josh:</u>

We know that v_{o}=48ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+48t+40       Eqn. (4).

<em><u>Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at </u></em>h(t)=0<em><u>. Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one. </u></em>

Thus we have for Ben, Eqn. (3) gives:

h(t)=0-16t^2+60t+40=0

Using the quadratic expression to find the roots of the quadratic we have:

t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec

Since time can only be positive we reject the t_{2} solution and we keep that Ben's book took t=4.3276 seconds to reach the ground.

Similarly solving for Josh we obtain

t_{1}=3.6794sec\\t_{2}=-0.6794sec

Thus again we reject the negative and keep the positive solution, so Josh's book took t=3.6794 seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482

So to answer the original question:

<em>Whose textbook reaches the ground first and by how many seconds?</em>

  • Josh's textbook reached the ground first
  • Josh's textbook reached the ground first by a difference of t=0.6482

3 0
3 years ago
What is the product of 14560 times 10
Andrews [41]
The product of 14560 and 10 is 145600
3 0
3 years ago
On Sunday, a local hamburger shop sold a combined total of 378 hamburgers and cheeseburgers. The number of cheeseburgers was sol
ipn [44]

Answer:

The number of hamburgers sold was 126.

Step-by-step explanation:

Let h = number of hamburgers sold

let c = cheeseburgers sold

c+h = 378

c = 2h

Substitute c =2h into c+h =278

c+h =378

2h+h = 378

Combine like terms

3h = 378

Divide by 3

3h/3 =378/3

h =126

The number of hamburgers sold was 126.

5 0
3 years ago
Please help need answers
bogdanovich [222]

see explanation below

(1) \frac{1}{5} × \frac{2}{2} = \frac{2}{10} = 0.2

(2) \frac{6}{25} × \frac{4}{4} = \frac{24}{100} = 0.24

(3) 2 \frac{3}{4} = 2 +\frac{75}{100} = 2.75

(4) 3 \frac{9}{10} = 3 + 0.9 = 3.9

(5) 1.25 = 1 \frac{1}{4} = \frac{5}{4}

(6) 3.29 = 3 \frac{29}{100} = \frac{329}{100}

(7) 0.65 = \frac{65}{100} = \frac{13}{20} in simplest form

(8) 5.6 = 5 \frac{6}{10} = 5 \frac{3}{5} = \frac{28}{5}

(9) he is incorrect

\frac{3}{5} × \frac{20}{20} = \frac{60}{100} = 0.6 ≠ 3.5


7 0
3 years ago
If x=3 and y=-2 then<br> x2 - xy<br> x²+2y
Naddika [18.5K]

Answer:

Step-by-step explanation:

(3)*2=6

-

(3)*(-2)=-6

6-(-6)=12

3*3=9

+2(-2)=-4

9+ -4= 5

4 0
3 years ago
Read 2 more answers
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