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4 years ago

15

Graph x≥−1, plz amswer. from khan

1 answer:

Mrrafil [7]4 years ago

7 0

Draw a number line. Plot -1 on the number line and make this a filled in circle. Shade to the right of the filled in circle to indicate the set of values x such that $x \ge -1$

So x could be -1 or it could be larger. The fact that x = -1 is possible is why we have a closed or filled in circle (instead of an open hole) at this endpoint.

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ABCDE and FGHJE are similar pentagons. If the perimeter of FGHJE is 7.3, what is the perimeter of ABCDE?

monitta

Polygons are said to be equal if the ratio of their corresponding sides and corresponding angles are equal. If the lengths are proportional, then it must be that the perimeter as well are proportional. Therefore, the correct answer is the second option. The perimeter of <span>ABCDE 14.6 units.</span>

3 0

3 years ago

Read 2 more answers

What is the error in -3.7 +-0.25= - .62

andrew11 [14]

Here is your answer they forgot to add the 3 in front this is your answer: -3.7+-0.25=-3.95 so as I'm look they added wrong its most like the 25 and the 7 were added wrong.

6 0

3 years ago

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Rewrite the expression using the GCF and the distributive property (39+91)

inessss [21]

Answer:

1. $13\cdot (3+7)$

2. $2,520

Step-by-step explanation:

1. Given numbers 39 and 91. Factorize them:

$39=3\cdot 13\\ \\91=7\cdot 13,$

then

$GCF(39,91)=13$

Use distributive property:

$39+91=3\cdot 13+7\cdot 13=13\cdot (3+7)$

2. Month salary = $2,100

To find 10% of $2,100, just multiply $2,100 by 0.1 (because 10% is 0.1 as decimal):

$\$2,100\cdot 0.1=\$210$

$210 is the monthly tax, then the year tax is

$\$210\cdot 12=\$2,520$

8 0

3 years ago

Consider the following equation. f(x, y) = y3/x, P(1, 2), u = 1 3 2i + 5 j (a) Find the gradient of f. ∇f(x, y) = Correct: Your

BaLLatris [955]

$f(x,y)=\dfrac{y^3}x$

a. The gradient is

$\nabla f(x,y)=\dfrac{\partial f}{\partial x}\,\vec\imath+\dfrac{\partial f}{\partial y}\,\vec\jmath$

$\boxed{\nabla f(x,y)=-\dfrac{y^3}{x^2}\,\vec\imath+\dfrac{3y^2}x\,\vec\jmath}$

b. The gradient at point P(1, 2) is

$\boxed{\nabla f(1,2)=-8\,\vec\imath+12\,\vec\jmath}$

c. The derivative of $f$ at P in the direction of $\vec u$ is

$D_{\vec u}f(1,2)=\nabla f(1,2)\cdot\dfrac{\vec u}{\|\vec u\|}$

It looks like

$\vec u=\dfrac{13}2\,\vec\imath+5\,\vec\jmath$

so that

$\|\vec u\|=\sqrt{\left(\dfrac{13}2\right)^2+5^2}=\dfrac{\sqrt{269}}2$

Then

$D_{\vec u}f(1,2)=\dfrac{\left(-8\,\vec\imath+12\,\vec\jmath\right)\cdot\left(\frac{13}2\,\vec\imath+5\,\vec\jmath\right)}{\frac{\sqrt{269}}2}$

$\boxed{D_{\vec u}f(1,2)=\dfrac{16}{\sqrt{269}}}$

7 0

4 years ago

What is the positive solution to the equation 0 = -x2 + 2x + 1?

katrin2010 [14]

Answer:

there is no solution

Step-by-step explanation:

1st switch sides: -x2+2x+1=0

2nd add similar elements: -2x + 2x = 0

1 = 0

Thus, this equation has no solution

hope i helped :)

5 0

4 years ago

Read 2 more answers