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3 years ago

Solve the following inequality. -21> -7c

Mathematics

2 answers:

valentinak56 [21]3 years ago

8 0

-7 because the number is closest to a positive number

BigorU [14]3 years ago

6 0

Divide both sides by -7, and flip the sign: 3 > c

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Susan subscribed to a video game rental service. She pays the same amount for each video game she rents plus a monthly fee to su

podryga [215]

Answer:

C-$6.00 monthly fee and $1.25 per video game

Step-by-step explanation:

7 0

3 years ago

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40 + (12) (27) + 18=

antiseptic1488 [7]

Answer:

Hey there!

40+12(27)+18

40+324+18

364+18

382

Let me know if this helps :)

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3 years ago

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Which pairs of quadrilaterals can be shown to be congruent using

Zielflug [23.3K]

Rigid transformation do not alter the sides and angle measurements of a shape.

The true statements are:

quadrilateral 1 and quadrilateral 2 - Not congruent
quadrilateral 1 and quadrilateral 3 - Not congruent
quadrilateral 1 and quadrilateral 4 - Not congruent
quadrilateral 2 and quadrilateral 3 - Congruent

From the graph, we have the following highlights

Quadrilaterals 2, 3 and 4 are congruent
Quadrilateral 1 does not have equal measure with any of the other quadrilaterals.

Hence, the first three statements are not congruent, while the last statement is congruent.

Read more about congruent shapes at:

brainly.com/question/24430586

3 0

2 years ago

Will give brainliest and extra points omg help ! /:

djyliett [7]

The answer is D i think

6 0

3 years ago

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Find the center that eliminates the linear terms in the translation of 4x^2 - y^2 + 24x + 4y + 28 = 0.(-3, 2)(-3,- 2)(4, 0)

baherus [9]

Step 1

Given;

$4x^2-y^2+24x+4y+28=0$

Required; To find the center that eliminates the linear terms

Step 2

$\begin{gathered} 4x^2-y^2+24x+4y=-28 \\ 4x^2+24x-y^2+4y=-28 \\ Complete\text{ the square }; \\ 4x^2+24x \\ \text{use the form ax}^2+bx\text{ +c} \\ \text{where} \\ a=4 \\ b=24 \\ c=0 \end{gathered}$ $\begin{gathered} consider\text{ the vertex }form\text{ of a }parabola \\ a(x+d)^2+e \\ d=\frac{b}{2a} \\ d=\frac{24}{2\times4} \\ d=\frac{24}{8} \\ d=3 \end{gathered}$ $\begin{gathered} Find\text{ the value of e using }e=c-\frac{b^2}{4a} \\ e=0-\frac{24^2}{4\times4} \\ e=0-\frac{576}{16}=-36 \end{gathered}$

Step 3

Substitute a,d,e into the vertex form

$\begin{gathered} a(x+d)^2+e \\ 4(x+_{}3)^2-36 \end{gathered}$ $\begin{gathered} 4(x+3)^2-36-y^2+4y=-28 \\ 4(x+3)^2-y^2+4y=\text{ -28+36} \\ \\ \end{gathered}$

Step 4

Completing the square for -y²+4y

$\begin{gathered} \text{use the form ax}^2+bx\text{ +c} \\ \text{where} \\ a=-1 \\ b=4 \\ c=0 \end{gathered}$ $\begin{gathered} consider\text{ the vertex }form\text{ of a }parabola \\ a(x+d)^2+e \\ d=\frac{b}{2a} \\ d=\text{ }\frac{4}{2\times-1} \\ d=\frac{4}{-2} \\ d=-2 \end{gathered}$ $\begin{gathered} Find\text{ the value of e using }e=c-\frac{b^2}{4a} \\ e=0-\frac{4^2}{4\times(-1)} \\ \\ e=0-\frac{16}{-4} \\ e=4 \end{gathered}$

Step 5

Substitute a,d,e into the vertex form

$\begin{gathered} a(y+d)^2+e \\ =-1(y+(-2))^2+4 \\ =-(y-2)^2+4 \end{gathered}$

Step 6

$\begin{gathered} 4(x+3)^2-y^2+4y=\text{ -28+36} \\ 4(x+3)^2-(y-2)^2+4=-28+36 \\ 4(x+3)^2-(y-2)^2=-28+36-4 \\ 4(x+3)^2-(y-2)^2=4 \\ \frac{4(x+3)^2}{4}-\frac{(y-2)^2}{4}=\frac{4}{4} \\ (x+3)^2-\frac{(y-2)^2}{2^2}=1 \end{gathered}$

Step 7

$\begin{gathered} \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1 \\ \text{This is the }form\text{ of a hyperbola.} \\ \text{From here } \\ a=1 \\ b=2 \\ k=2 \\ h=-3 \end{gathered}$

Hence the answer is (-3,2)

4 0

1 year ago