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3 years ago

A formula can be rewritten to isolate a different _______.

Mathematics

1 answer:

ahrayia [7]3 years ago

5 0

Answer:

Variables

Step-by-step explanation:

EDG2021

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Find the complement of the set given that U = {x | x ∈ I and −3 ≤ x ≤ 7}. (Enter your answers as a comma-separated list.) {x | x

kicyunya [14]

Huh?? Can u attach a picture or some

7 0

3 years ago

(3x5 y2)(6x3y7) simplify

vladimir2022 [97]

Answer:

(1): "x7" was replaced by "x^7". 1 more similar replacement(s).

Equation at the end of step 1

(0 - (3 • (x5))) • (2•3x7)

Equation at the end

(0 - 3x5) • (2•3x7)

Multiplying exponential expressions

3.1 x5 multiplied by x7 = x(5 + 7) = x12

Multiplying exponents:

3.2 31 multiplied by 31 = 3(1 + 1) = 32

Final result :

( -32•2x12)

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Find the coordinates of the intersection of the diagonals of parallelogram HJKL with the given vertices: H(-1, 4), J(3, 3), K(3,

lakkis [162]

Answer:

(1, 1)

Step-by-step explanation:

Given vertices of the parallelogram:

H = (-1, 4)
J = (3, 3)
K = (3, -2)
L = (-1, -1)

Therefore the parallel sides are:

$\sf \overline{HJ} \parallel \overline{LK}\:\: \textsf{ and }\:\: \overline{LK} \parallel \overline{HL}$

Therefore, the diagonals of the parallelogram are:

$\sf \overline{LJ} \:\: \textsf{ and }\:\:\overline{HK}$

To find the coordinates of the intersection of the diagonals, either:

draw a diagram (see attached) and determine the point of intersection of the diagonals from the diagram, or
determine the midpoint of either diagonal (as the diagonals of a parallelogram bisect each other, i.e. divide into 2 equal parts).

Midpoint between two points

$\textsf{Midpoint}=\left(\dfrac{x_2+x_1}{2},\dfrac{y_2+y_1}{2}\right)\quad \textsf{where}\:(x_1,y_1)\:\textsf{and}\:(x_2,y_2)\:\textsf{are the endpoints}}\right)$

To find the midpoint of diagonal LJ, define the endpoints:

$(x_1,y_1)=L=(-1, -1)$
$(x_2,y_2)=J=(3,3)$

Substitute the defined endpoints into the formula and solve:

$\begin{aligned} \implies \textsf{Midpoint of LJ} & =\left(\dfrac{3-1}{2},\dfrac{3-1}{2}\right)\\ & =\left(\dfrac{2}{2},\dfrac{2}{2}\right)\\ & =\left(1,1\right) \end{aligned}$