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3 years ago

EXPLAIN A LOT OF POINTS!!!

Mathematics

1 answer:

IrinaVladis [17]3 years ago

3 0

Answer:

x−(−16)=−28+12

x+16=−28+12

x+16=(−28+12)

x+16=−16

x+16−16=−16−16

x=−32

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the total cost, c, of an item purchased from a store is c = p .07p, where p is the price before sales tax, and there is a 7% sal

alekssr [168]

C = p + 0.07p
c = p(1 + 0.07)
c = 1.07p

6 0

3 years ago

Find the restricted values of x for the following rational expression. If there are no restricted values of x, indicate "No Rest

melamori03 [73]

Given rational expression is

$-\frac{8x}{8x^2+2x}$

Now we need to find the restricted values if any for this rational expression.

Restricted values means the possible values of the used variable (x) that will make denominator 0 as division by 0 is not defined.

So to find the restricted values, we just set denominator equal to 0 and solve for x

$8x^2+2x=0$

$2x(4x+1)=0$

2x=0 or 4x+1=0

x=0 or 4x=-1

x=0 or x=-1/4

Hence final answer is x=0, -1/4

3 0

3 years ago

11. find the value of x in the kite below. E (6x + 6) F Р G

cluponka [151]

14.

Step-by-step explanation:

6x+6=90 (being a straight line)

6x=90-6

6x=84

x= 84/6

x=14.

8 0

3 years ago

Help Please I’ll Give Brainliest :)

Mnenie [13.5K]

Answer:

5 tables

Step-by-step explanation:

3 = 24

1 = 24 ÷ 3 = 8

40 chairs = 40 ÷ 8 = 5

Hope this helps!

7 0

3 years ago

Read 2 more answers

Determine whether the set of vectors <img src="https://tex.z-dn.net/?f=%20v_%7B1%3D%283%2C2%2C1%29%2C%20v_%7B2%7D%20%3D%28-1%2C-

Korolek [52]

Since each vector is a member of $\mathbb R^3$ , the vectors will span $\mathbb R^3$ if they form a basis for $\mathbb R^3$ , which requires that they be linearly independent of one another.

To show this, you have to establish that the only linear combination of the three vectors $c_1\mathbf v_1+c_2\mathbf v_2+c_3\mathbf v_3$ that gives the zero vector $\mathbf0$ occurs for scalars $c_1=c_2=c_3=0$ .

$c_1\begin{bmatrix}3\\2\\1\end{bmatrix}+c_2\begin{bmatrix}-1\\-2\\-4\end{bmatrix}+c_3\begin{bmatrix}1\\1\\-1\end{bmatrix}=(0,0,0)\iff\begin{bmatrix}3&-1&1\\2&-2&1\\1&-4&-1\end{bmatrix}\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$

Solving this, you'll find that $c_1=c_2=c_3=0$ , so the vectors are indeed linearly independent, thus forming a basis for $\mathbb R^3$ and therefore they must span $\mathbb R^3$ .

4 0

3 years ago