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2 years ago

Math unit rate can someone please help me

Mathematics

1 answer:

Finger [1]2 years ago

7 0

Answer: All I know is C is one of them. I’m trying lol

Step-by-step explanation:

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Philip saves $8 each month. How many months will it take him to save at least $60?

mars1129 [50]

7.5 months for him to save up to 60 ;)

8 0

3 years ago

Read 2 more answers

Are the triangle below similar

sweet [91]

Answer:

Yes ,because of SAS similarity theorem

Step-by-step explanation:

In ΔYOM , ΔTON

∠YOM = ∠TON

$\frac{OY}{OT}=\frac{96}{84}=\frac{8}{7}\\\\\frac{OM}{ON}=\frac{104}{91}=\frac{8}{7}$

SAS similarity theorem

4 0

2 years ago

Read 2 more answers

Calculating the Surface Area of a Triangular Prism The triangular prism shown has 1235 triangular face(s) and 1235 lateral face(

GenaCL600 [577]

Corrected Question

The triangular prism shown has 1, 2, 3 or 5 triangular faces and 1,2,3, or 5 lateral faces.

The area of one triangular face is 7.5,8.125,15,or 17 $mm^2$

The surface area of the triangular prism is 55.5,127.5,135,or 150 $mm^2$

Answer:

(B)2 triangular faces

(A)Area of one Triangular Face $=7.5mm^2$

(C)Total Surface Area of the Triangular prism $=135mm^2$

Step-by-step explanation:

The triangular prism is attached.

The triangular prism shown has 2 triangular faces and 3 lateral faces.

Base of the triangle =2.5mm

Height of the Triangle =6mm

Area of one Triangular Face $=\frac{1}{2}*6*2.5=7.5mm^2$

The dimensions of the lateral rectangles are:

2.5 mm by 8mm
6 mm by 8mm
6.5 mm by 8mm

Therefore, total surface area of the triangular prism

=2(Area of one Triangular Face)+Area of 3 rectangular faces

$=2(7.5)+ (2.5 X 8+ 6 X 8 + 6.5 X 8)\\=15+120\\=135mm^2$

Total Surface Area of the Triangular prism $=135mm^2$

8 0

2 years ago

What is the value of each expression?

Hunter-Best [27]

Give me points thanks

6 0

3 years ago

Urgent!

boyakko [2]

Step-by-step explanation:

18.

the last one.

-4.9t² + 24.5t + 117.6 = 0

the ball starts at 117.6 meters height. then, during the first seconds the thrust upwards is adding height, until finally, with more and more seconds passing, gravity will win with a vengeance and pulls the ball down faster and faster than any other force in any other direction.

19.

the general solution for a quadratic equation

y = ax² + bx + c

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

x = t

a = -4.9

b = 24.5

c = 117.6

t = (-24.5 ± sqrt(600.25 - -4×4.9×117.6))/-9.8 =

= (-24.5 ± sqrt(600.25 + 2304.96))/-9.8 =

= (-24.5 ± sqrt(2905.21))/-9.8 = (-24.5 ± 53.9)/-9.8

t1 = (-24.5 + 53.9)/-9.8 = -3

t2 = (-24.5 - 53.9)/-9.8 = 8

a negative time does not make any sense, so the real solution is : the ball reached the ground after 8 seconds.

20.

now the equation is

-4.9t² + 24.5t + 117.6 = 49

but this is quickly transformed again into a "= 0" equation :

-4.9t² + 24.5t + 68.6 = 0

t = (-24.5 ± sqrt(600.25 - -4×4.9×68.6))/-9.8 =

= (-24.5 ± sqrt(600.25 + 1344.56))/-9.8 =

= (-24.5 ± sqrt(1944.81))/-9.8 = (-24.5 ± 44.1)/-9.8

t1 = (-24.5 + 44.1)/-9.8 = -2

t2 = (-24.5 - 44.1)/-9.8 = 7

again, negative time does not make sense.

the ball will be at 49 m height after 7 seconds.

5 0

2 years ago