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3 years ago

For the equation 3(7 + 2) = 3x + k, what value of k will create an equation with infinitely many solutions?

Mathematics

2 answers:

Natasha2012 [34]3 years ago

4 0

Answer:

ygftyhhhhgffyujhhgyjjjhhguj

Marta_Voda [28]3 years ago

3 0

This would be C.

Mark me brainlest if this work!!

You might be interested in

What temperature is 10 degrees less then 6 degrees celsius

choli [55]

F=(9*6/5)+32

F=54/5+32

10.8+32

42.8

42.8-10=32.8

4 0

3 years ago

A teacher wants to give 3 markers to

Novay_Z [31]

First, find out how many markers in total the teacher will need.
$25*3=75$
Then, find out how many packages the teacher will need by dividing by 8 because that's how many are in each package.
$\frac{75}{8}=9.375$
Now because there is a remainder of .375 that tells us that there are not enough markers in 9 packages to give all students 3 each.

She will have to buy an additional package to give every student 3 markers each.

10 Packages are needed.

8 0

3 years ago

Read 2 more answers

What is 15% of 60? Create a model to prove your answer.

ICE Princess25 [194]

Answer:

Step-by-step explanation:

15 is 1/4 of 60, and 1/4 is 25%. So, 15 is 25% of 60.

4 0

4 years ago

Read 2 more answers

Let vector F = (6 x^2 y + 2 y^3 + 4 e^x) i + (7 e^{y^2} + 54 x) j . Consider the line integral of vector F around the circle of

balu736 [363]

Denote the circle of radius $a$ by $C$ . $C$ is simple and closed, so by Green's theorem the line integral reduces to a double integral over the interior of $C$ (call it $D$ ):

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_C(6x^2y+2y^3+4e^x)\,\mathrm dx+(7e^{y^2}+54x)\,\mathrm dy$

$=\displaystyle\iint_D\left(\frac{\partial(7e^{y^2}+54x)}{\partial x}-\frac{\partial(6x^2y+2y^3+4e^x)}{\partial y}\right)\,\mathrm dx\,\mathrm dy$

$=\displaystyle\iint_D(54-6x^2-6y^2)\,\mathrm dx\,\mathrm dy$

$D$ is a circle of radius $a$ , so we can write the double integral in polar coordinates as

$\displaystyle\iint_D(54-6x^2-6y^2)\,\mathrm dx\,\mathrm dy=\int_0^{2\pi}\int_0^a(54-6r^2)r\,\mathrm dr\,\mathrm d\theta$

a. For $a=1$ , we have

$\displaystyle\int_0^{2\pi}\int_0^1(54-6r^2)r\,\mathrm dr\,\mathrm d\theta=2\pi\int_0^1(54r-6r^3)\,\mathrm dr=\boxed{51\pi}$

b. Let $I(a)$ denote the integral with unknown parameter $a$ ,

$I(a)=12\pi\int_0^a(9r-r^3)\,\mathrm dr\,\mathrm d\theta$

By the fundamental theorem of calculus,

$I'(a)=12\pi(9a-a^3)$

$I(a)$ has critical points when

$12\pi(9a-a^3)=12\pi a(9-a^2)=0\implies a=0,a=\pm3$

If $a=0$ , then line integral is 0, so we ignore that critical point. For the other two, we would find $I(\pm3)=243\pi$ .

8 0

3 years ago

A normal distribution is symmetric; therefore the probability of being below the mean is 0.50 and the probability of being above

ladessa [460]

Answer:

a. True

Step-by-step explanation:

We have been given a statement. We are asked to determine whether the given statement is true or false.

Statement: A normal distribution is symmetric; therefore the probability of being below the mean is 0.50 and the probability of being above the mean is 0.50.

We know that normal distribution is continuous probability distribution.

We know that the total area under a normal distribution curve is equal to 1. We also know that mean represents center of the normal distribution curve.

The half area will be below center and half area will be above center.

Since 0.50 is half of 1, therefore, the probability of being below the mean is 0.50 and the probability of being above the mean is 0.50 and given statement is true.

5 0

4 years ago