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joja [24]
3 years ago
13

Please help me with this one

Mathematics
1 answer:
Alexxx [7]3 years ago
7 0
4 (points for an A) x 21 (As) = 84
3 (points for a B) x 24 (Bs) =72
2 (points for a C) x 15 (cs) = 30
84 + 72 + 30 = 186 (total points)
186 divided by 60 (total credits) = 3.1
3.1 GPA
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Consider the enlargement of the rectangle. Use the proportion to find the missing dimension bod the original rectangle. (Sorry t
fgiga [73]

Here's the right way:

\dfrac{x}{\frac 3 5} = \dfrac{20}{12}

x = \dfrac{20}{12} \cdot \dfrac{3}{5} = 1

Here's the way they want you to do it:

\dfrac{x}{\frac 3 5} = \dfrac{20}{12}

12 x = 20 \times \dfrac{3}{5} = 12 \textrm{ FIRST BOX}

x = 12/12 = 1 \textrm{ SECOND BOX}

6 0
3 years ago
Read 2 more answers
Solve equations by using elimination <br> -4+5y= 13<br> X-5y=21<br> ( , )
tigry1 [53]

Answer:

y=17/5 ,x= 38

Step-by-step explanation:

-4+5y =13 add 4 both side

4-4+5y=13+4

5y=17

5y/5=17/5 y=17/5

x-5y=21

x-5(17/5)=21

x-17=21

x=21+17= 38

5 0
3 years ago
What does 17/10x 10/17 equal
Rasek [7]
\frac{17}{10}\times\frac{10}{17}=\frac{17\times10}{10\times17}=\frac{170}{170}=1

5 0
1 year ago
Evaluate the expression when m=-6.<br> 2<br> m + 5m - 4
Brums [2.3K]

Answer:

2

Step-by-step explanation:

m^2 + 5m - 4

Let m = -6

( -6) ^2 + 5(-6) -4

Exponents first

36 + 5(-6) -4

Then multiply

36 -30 -4

Then subtract

6 -4

2

5 0
3 years ago
The height of a stuntperson jumping off a building that is 20 m high is modeled by the equation h = 20 -57, where t is the time
cupoosta [38]

A stuntman jumping off a 20-m-high building is modeled by the equation h = 20 – 5t2, where t is the time in seconds. A high-speed camera is ready to film him between 15 m and 10 m above the ground. For which interval of time should the camera film him?

Answer:

1\leq t\geq \sqrt{2}

Step-by-step explanation:

Given:

A stuntman jumping off a 20-m-high building is modeled by the equation

h =20-5t^{2}-----------(1)

A high-speed camera is ready to making film between 15 m and 10 m above the ground

when the stuntman is 15m above the ground.

height h = 15m  

Put height value in equation 1

15 =20-5t^{2}

5t^{2} =20-15

5t^{2} =5

t^{2} =1

t =\pm1

We know that the time is always positive, therefore t=1

when the stuntman is 10m above the ground.

height h = 10m  

Put height value in equation 1

10 =20-5t^{2}

5t^{2} =20-10

5t^{2} =10

t^{2} =\frac{10}{5}

t^{2} =2

t=\pm\sqrt{2}

t=\sqrt{2}

Therefore ,time interval of camera film him is 1\leq t\geq \sqrt{2}

7 0
3 years ago
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