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3 years ago

Solve for c.

Mathematics

2 answers:

sveticcg [70]3 years ago

5 0

Answer: the correct option is

(D) $c=\dfrac{d+ab}{a}.$

Step-by-step explanation: We are given to solve the following equation for the value of c :

$a(c-b)=d~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

To solve the given equation for c, we need to take c on one side and all other values on the other side of the equation.

The solution for c of equation (i) is as follows :

$a(c-b)=d\\\\\Rightarrow ac-ab=d\\\\\Rightarrow ac=d+ab\\\\\Rightarrow c=\dfrac{d+ab}{a}.$

Thus, the required value of c is $c=\dfrac{d+ab}{a}.$

Option (D) is CORRECT.

ICE Princess25 [194]3 years ago

3 0

a(c−b)=d
ac - ab = d
ac = d + ab
c = (d + ab) / a

answer is
D. c = (d + ab) / a

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Pls help me figure out the slope

cluponka [151]

Answer:

$m=\frac{-4}{5}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Algebra I

Slope Formula: $m=\frac{y_2-y_1}{x_2-x_1}$

Step-by-step explanation:

Step 1: Define

Find points from graph.

Point (-3, 1)

Point (2, -3)

Step 2: Find slope m

Substitute: $m=\frac{-3-1}{2+3}$
Subtract/Add: $m=\frac{-4}{5}$

7 0

2 years ago

Need help with #1 #4 #7 plz giving 15 points

valentina_108 [34]

Answer:

Q1: p = - 33

Q2: d = - 99

Q3: t = - 13

Step-by-step explanation:

Q1: $$ \textbf{-} \frac{\textbf{p}}{\textbf{3}} \hspace{1mm} \textbf{-} \hspace{1mm} \textbf{8} \hspace{1mm} \textbf{=} \hspace{1mm} \textbf{3}$

We solve this taking LCM.

We get: $\frac{-p - 24}{3} = 3$

$\implies - p - 24 = 9$

$\implies \textbf{p} \hspace{1mm} \textbf{=} \hspace{1mm} \textbf{- 33}$

Q4: $\frac{\textbf{d}}{\textbf{11}} \hspace{1mm} \textbf{-} \hspace{1mm} \textbf{4} \hspace{1mm} \textbf{=} \hspace{1mm} \textbf{- 13}$

Again we proceed like Q1 by taking LCM.

We get: $\frac{d - 44}{11} = - 13$

$\implies d - 44 = - 13 \times 11 = - 143$

$\implies d = - 143 + 44$

$\implies \textbf{d} \hspace{1mm} \textbf{=} \hspace{1mm} \textbf{- 99}$

Q7: 5t + 12 = 4t - 1

We club the like terms on either side.

$\implies 5t - 4t = - 1 - 12$

$\implies (5 -4)t = - 13$

$\implies \textbf{t} \hspace{1mm} \textbf{=} \hspace{1mm} \textbf{- 13}$

Hence, the answer.

7 0

2 years ago

The area of a rectangle is 27 m^2 , and the length of the rectangle is 3 m less than twice the width. Find the dimensions of the

vagabundo [1.1K]

Let $L$ be the length of the rectangle and $W$ be the width. In the problem it is given that $L=2W-3$ . It is also given that the area $LW=27$ . Substituting in the length in terms of width, we have $W(2W-3)=27 \\ 2W^2-3W-27=0 \\ (2W-9)(W+3)=0$ . Using the zero product property, $2W-9=0 \text{ or } W+3=0$ . Solving these we get the width $W=4.5 \text{ or } -3$ . However, it doesn't make sense for the width to be negative, so the width must be $\boxed{4.5 \text{ m}}$ . From that we can tell the length $L=2(4.5)-3=\boxed{6 \text{ m}}$ .