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3 years ago

WILL GIVE BRAINLY PLEASE ANSWER THIS MATHEMATICAL QUESTION STEP BY STEP!!!!!!! (-5 − 2) × (-3)

Mathematics

2 answers:

Oksanka [162]3 years ago

7 0

Answer:

-5-2 = -7 Step 1 from abaolute value quality

-7*(-3) =21 step 2 from absolute value quality

KatRina [158]3 years ago

6 0

Answer:

Step-by-step explanation:

Start with parenthesis:

(-5 -3) x (-3)

-8 x -3

Then multiply:

-8 x -3 = 24

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Two increased by the product of a number and 9 is greater than or equal to 25.

Marianna [84]

Answer:

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Step-by-step explanation:

first we can write the inequality

2+9x≥25

we can now solve this with algebra

9x≥23

x≥2.56

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3 years ago

If a cross-country runner runs a 55 kilometer (km) race, how many miles is that? Round your answer to one decimal place.

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2 years ago

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2 years ago

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Write an equivalent expression to -3(6x^2-10x+9)+8x^2(3x+4)

pychu [463]

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Step-by-step explanation: hope this help :)

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2 years ago

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Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2/3 + y2/3 = 4 (−3 3 , 1)

vovikov84 [41]

Answer with Step-by-step explanation:

We are given that an equation of curve

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$

We have to find the equation of tangent line to the given curve at point $(-3\sqrt3,1)$

By using implicit differentiation, differentiate w.r.t x

$\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=0$

Using formula : $\frac{dx^n}{dx}=nx^{n-1}$

$\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=-\frac{2}{3}x^{-\frac{1}{3}}$

$\frac{dy}{dx}=\frac{-\frac{2}{3}x^{-\frac{1}{3}}}{\frac{2}{3}y^{-\frac{1}{3}}}$

$\frac{dy}{dx}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}$

Substitute the value x= $-3\sqrt3,y=1$

Then, we get

$\frac{dy}{dx}=-\frac{(-3\sqrt3)^{-\frac{1}{3}}}{1}$

$\frac{dy}{dx}=-(-3^{\frac{3}{2}})^{-\frac{1}{3}}=-\frac{1}{-(3)^{\frac{3}{2}\times \frac{1}{3}}}=\frac{1}{\sqrt3}$

Slope of tangent=m= $\frac{1}{\sqrt3}$

Equation of tangent line with slope m and passing through the point $(x_1,y_1)$ is given by

$y-y_1=m(x-x_1)$

Substitute the values then we get

The equation of tangent line is given by

$y-1=\frac{1}{\sqrt3}(x+3\sqrt3)$

$y-1=\frac{x}{\sqrt3}+3$

$y=\frac{x}{\sqrt3}+3+1$

$y=\frac{x}{\sqrt3}+4$

This is required equation of tangent line to the given curve at given point.

8 0

3 years ago