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NeX [460]
3 years ago
15

What is ten times two?

Mathematics
1 answer:
Anton [14]3 years ago
6 0
Ten times two is twenty
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Express 42 base 7 in base 10​
xxTIMURxx [149]

Answer:

correct 42 in bsae 10 is 60

8 0
3 years ago
What is the area of a regular nonagon with 10 cm sides and an apothem if 6cm?
bearhunter [10]

Answer:

The area of the nonagon is 270cm²

Step-by-step explanation:

The area of a regular polygon is given by

=  \frac{1}{2} ap

where 'a' is the apothem and 'p' is the perimeter.

The apothem is a=6cm.

The perimeter is p=9×10=90cm.

We substitute the apothem and the perimeter to get:

=  \frac{1}{2}  \times 6 \times 90 {cm}^{2}

= 270 {cm}^{2}

8 0
3 years ago
What is the slope of the graph shown below?
GenaCL600 [577]

Answer:

B: -2

Step-by-step explanation:

Find the slope by counting how many times it would take to reach a point. For example, (0,1). Or use you could use the slope formula by finding two coordinates

3 0
3 years ago
Find all values of x (if any) where the tangent line to the graph of the given equation is horizontal. HINT [The tangent line is
Nostrana [21]
By "y = −9x2 − 2x" I assume you meant  <span>y = −9x^2 − 2x (the "^" symbol represents exponentiation).

Let's find the first derivative of y with respect to x:  dy/dx = -18x - 2.  This is equivalent to the slope of the tangent line to the (parabolic) curve.  Now let this derivative (slope) = 0 and solve for the critical value:  -18x - 2 = 0, or
-18x = 2.  Solving for x,   x = -2/18,    or    x = -1/9.

When x = -1/9, y = -9(-1/9)^2 - 2(-1/9).  This simplifies to y = -9/9 + 2/9, or 
y = -7/9.

The only point at which the tangent to the curve is horiz. is (-1/9,-7/9).</span>
6 0
3 years ago
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
4 years ago
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