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3 years ago

Need help pls with this problem

Mathematics

1 answer:

tensa zangetsu [6.8K]3 years ago

8 0

Answer:

100

Step-by-step explanation:

The statement Tell us: a=2 b=5

We have

(a*b)^2=(2*5)^2=(10)^2

Finally we have:

10*10=100

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The camp director , Mrs Marshall, put all 1,265 backpacks on the bus. Each backpack weighed seven pounds. How many pounds did Mr

posledela

Answer: 8855 pounds

Step-by-step explanation: Simply multiply 1265 x 7

6 0

3 years ago

Read 2 more answers

Factor 200t+8t^3-80t^2

valina [46]

Answer:

8t(t - 5)^2

Step-by-step explanation:

I would immediately rearrange these terms in descending order by powers of t:

8t^3 - 80t^2 + 200t

Then I'd factor out 8t:

8t(t^2 - 10 t + 25).

Finally, I'd factor the quadratic t^2 - 10 t + 25:

(t - 5)^2

and so 200t+8t^3-80t^2 in factored form is 8t(t - 5)^2.

6 0

3 years ago

Read 2 more answers

Which expression is not equivalent to the other three?

vagabundo [1.1K]

Answer:

b. 4+7x-2

Explanation:

when you add like terms for the first equation it becomes 7x-6

the second one becomes 7x+2

the third one becomes 7x-6

and the last one is also 7x-6

which means the second one is not equivalent

7 0

3 years ago

I really need help I’ve worked it out already though

Vikki [24]

<h2>Answer:</h2><h3>75(5r)+75(3d)+75(4f)</h3>

Solution,

Cost of roses in 1 bouquet=5r

Cost of daisies in 1 bouquet=3d

cost of green filler in 1 bouquet=4f

Total no. of bouquet=75

cost of one bouquet=5r+3d+4f

total=75(5r+3d+4f)

 =75(5r)+75(3d)+75(4f)

hope it helps

Good luck on your assignment

8 0

3 years ago

3. A person standing 3 feet away from the base of a lamp post casts a shadow 4 feet long. If the person is 6 feet tall and walks

german

Answer:

533.33 ft/min

Step-by-step explanation:

We are given that

x=3 feet

y=4 feet

Height of person,h=6 feet

$\frac{dx}{dt}=400 ft/min$

Triangle ABC and A'B'C are similar because all right triangles are similar

By using similarity property

$\frac{h'}{6}=\frac{7}{4}$

$h'=10.5 feet$

$x+y=3+4=7 feet$

$\frac{h'}{6}=\frac{x+y}{y}$

$\frac{10.5}{6}=\frac{x+y}{y}$

$10.5y=6x+6y$

$10.5y-6y=6x$

$4.5y=6x$

Differentiate w.r.t t

$4.5\frac{dy}{dt}=6\frac{dx}{dt}$

$\frac{dy}{dt}=\frac{6\times 400}{4.5}ft/min$

$\frac{dy}{dt}=533.33ft/min$

Hence, If the person is 6 feet tall and walks away from the lamp post at a speed of 400 feet per minute then his shadow moving at the rate 533.33 ft/min

7 0

3 years ago