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I have done a and b but I am really struggling on C

Rashid [163]

Hmmm, just follow what ur gut thinks ifthat make sense

3 0

3 years ago

How do you write 78,621,087 in expanded notation and expanded form?

Inga [223]

70,000,000+8,000,000+600,000+20,000+1,000+80+7 = 78,621,087

4 0

3 years ago

the length of the sides of a square are initially 0 cm and increase at a constant rate of 11 cm per second. suppose the function

Likurg_2 [28]

The function of the area of the square is A(t)=121 $t^{2}$

Given that The length of a square's sides begins at 0 cm and increases at a constant rate of 11 cm per second. Assume the function f determines the area of the square (in cm2) given several seconds, t since the square began growing and asked to find the function of the area

Lets assume the length of side of square is x

$\frac{dx}{dt} =$ 11 $\frac{cm}{sec}$

⇒x=11t

Area of square= $(length of side)^{2}$

Area of square= $(11t)^{2}$ {as the length of side is 11t}{varies by time}

Area of square=121 $t^{2}$

Therefore,The function of the area of the square is A(t)=121 $t^{2}$

Learn more about The function of the area of the square is A(t)=121 $t^{2}$

Given that The length of a square's sides begins at 0 cm and increases at a constant rate of 11 cm per second. Assume the function f determines the area of the square (in cm2) given several seconds, t since the square began growing and asked to find the function of the area

Lets assume the length of side of square is x

$\frac{dx}{dt} =$ 11 $\frac{cm}{sec}$

⇒x=11t

Area of square= $(length of side)^{2}$

Area of square= $(11t)^{2}$ {as the length of side is 11t}{varies by time}

Area of square=121 $t^{2}$

Therefore,The function of the area of the square is A(t)=121 $t^{2}$

Learn more about area here:

brainly.com/question/27683633

#SPJ4

3 0

2 years ago

No biggy just need some help

Temka [501]

Answer:

betwen year 5 and 6

Step-by-step explanation:

th prize will drop betwen 5 and 6 years.

i think its the correct answer

8 0

2 years ago

Express the trig ratios as fractions in simplest terms.

Neporo4naja [7]

Answer:

See below

Step-by-step explanation:

$\cos M=\frac{adjacent}{hypotenuse}=\frac{56}{70}=\frac{4}{5}\\\\\sin L=\frac{opposite}{hypotenuse}=\frac{56}{70}=\frac{4}{5}$

Hence, both ratios are equal to each other

3 0

2 years ago