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3 years ago

F (t) = 2t - 5 find f(7)

Mathematics

2 answers:

jeyben [28]3 years ago

7 0

Answer:9

Step-by-step explanation:2(7)-5=9

anzhelika [568]3 years ago

6 0

Answer:

explanation=

2(7)-5

14-5

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Tara wants to weigh her three stuffed animals. They will only fit on the scale two at a time. Together Addie and Missy weight 18

Pepsi [2]

Answer: Addie weighs 4 ounces

Missy weighs 14 ounces

Corky weighs 8 ounces

Step-by-step explanation:

Let a represent the weight of Addie.

Let m represent the weight of Missy.

Let c represent the weight of Corky.

Together Addie and Missy weigh 18 ounces. This means that

a + m = 18 - - - - - - - - - 1

Missy and Corky weigh 22 ounces. This means that

m + c = 22

m = 22 - c - - - - - - - - - - 2

Addie and Corky weigh 12 ounces. This means that

a + c = 12

a = 12 - c - - - - - - - - - - - 3

Substituting equation 2 and equation 3 into equation 1, it becomes

22 - c + 12 - c = 18

34 - 2c = 18

- 2c = 18 - 34 = - 16

c = - 16/ - 2 = 8

Substituting c = 8 into equation 2, it becomes

m = 22 - 8

m = 14

Substituting c = 8 into equation 3, it becomes

a = 12 - 8

a = 4

3 0

3 years ago

How to do this and what's the answer

Darya [45]

The answer to this question is b I got it wrong With picking a and my teacher said it was b

6 0

3 years ago

Find dy/dx if y =x^3+5x+2/x²-1

stiks02 [169]

<u>Differentiate using the Quotient Rule</u> –

$\qquad$ $\pink{\twoheadrightarrow \sf \dfrac{d}{dx} \bigg[\dfrac{f(x)}{g(x)} \bigg]= \dfrac{ g(x)\:\dfrac{d}{dx}\bigg[f(x)\bigg] -f(x)\dfrac{d}{dx}\:\bigg[g(x)\bigg]}{g(x)^2}}\\$

According to the given question, we have –

f(x) = x^3+5x+2
g(x) = x^2-1

Let's solve it!

$\qquad$ $\green{\twoheadrightarrow \bf \dfrac{d}{dx}\bigg[ \dfrac{x^3+5x+2 }{x^2-1}\bigg]} \\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{(x^2-1) \dfrac{d}{dx}(x^3+5x+2) - ( x^3+5x+2) \dfrac{d}{dx}(x^2-1)}{(x^2-1)^2 }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{(x^2-1)(3x^2+5) - ( x^3+5x+2) 2x}{(x^2-1)^2 }\\$

$\qquad$ $\pink{\sf \because \dfrac{d}{dx} x^n = nx^{n-1} }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-(2x^4+10x^2+4x)}{(x^2-1)^2 }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-2x^4-10x^2-4x}{(x^2-1)^2 }\\$

$\qquad$ $\green{\twoheadrightarrow \bf \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}\\$

$\qquad$ $\pink{\therefore \bf{\green{\underline{\underline{\dfrac{d}{dx} \dfrac{x^3+5x+2 }{x^2-1}} = \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}}}}\\\\$

7 0

2 years ago

7. What is the value of x for the triangle drawn below?

slega [8]

Answer:

where is the triangle?

8 0

3 years ago

Consider the region bounded by

Montano1993 [528]

The equation is a circle centered at the origin with radius 8 (sqrt(64))
Therefore, the bounded region is just a quarter circle in the first quadrant.

Riemann Sum: ∑⁸ₓ₋₋₀(y²)Δx=∑⁸ₓ₋₋₀(64-x²)Δx

Definite Integral: ∫₀⁸(y²)dx=∫₀⁸(64-x²)dx

4 0

3 years ago