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MissTica
3 years ago
13

10x-35+3ax=5ax-7a solve for "a" for which the equation is an identity.

Mathematics
1 answer:
PilotLPTM [1.2K]3 years ago
6 0
10x-35+3ax=5ax-7a \\ 3ax-5ax+7a = -10x+35 \\ -2ax+7a = -10x+35 \\ a(-2x+7)=-10x+35 \\\\ a= \frac{-10x+35}{-2x+7} \\\\ a =  \frac{5(-2x+7)}{(-2x+7)} \\\\ a=5
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If ABC and EDC are similar, what is the value<br> of x?<br> A
abruzzese [7]

Answer:

x = 15

Step-by-step explanation:

Since the triangles are similar then the ratios of corresponding sides are equal, that is

BC/DC = EC/AC , substitute values

3x×5/32 - 5x-5/56 ( cross- multiply )

56(3x - 5) = 32(5x - 5) ← distribute and simplify both sides

168x - 280 = 160x - 160 ( subtract 160x from both sides )

8x - 280 = - 160 ( add 280 to both sides )

8x = 120 ( divide both sides by 8 )

x = 15

3 0
3 years ago
If a+b = -0.5 and ab = -2 , then what is 4a^3+9b?
m_a_m_a [10]

Answer:

pls be more cleare

Step-by-step explanation:

type the question properly and clearly

5 0
2 years ago
PLEASE HELP ANSWER !!!
geniusboy [140]

Answer:

the answer is 81

Step-by-step explanation:

1. find 75

2. add 75 + 24

3. subtract from 180 to find x

3 0
3 years ago
Situation:
FromTheMoon [43]

Answer:

5.6 days

Step-by-step explanation:

We are given;

Initial Mass; N_o = 25 g

Mass at time(t); N_t = 25/2 = 12.5 (I divide by 2 because we are dealing with half life)

k = 0.1229

Formula is given as;

N_t = N_o•e^(-kt)

Plugging in the relevant values;

12.5 = 25 × e^(-0.1229t)

e^(-0.1229t) = 12.5/25

e^(-0.1229t) = 0.5

(-0.1229t) = In 0.5

-0.1229t = -0.6931

t = -0.6931/-0.1229

t = 5.6 days

4 0
3 years ago
A software designer is mapping the streets for a new racing game. all of the streets are depicted as either perpendicular or par
ElenaW [278]

The equation of the central street PQ is  -1.5x - 3.5y = -31.5 option (b) is correct.

<h3>What is a straight line?</h3>

A straight line is a combination of endless points joined on both sides of the point.

We have a straight line:

-7x+3y=-21.5  

Convert it to the general form given below:

\rm y=mx+c

\rm 3y=-21.5+7x\\\\ \rm y = \frac{-21.5}{3}+\frac{7}{3}x or

\rm y = \frac{7}{3}x-\frac{21.5}{3}

m = \frac{7}{3}   (Slope of AB line)

For the slope(m') of the PQ line:

\rm m'=-\frac{1}{m}    ( because AB and PQ are perpendicular to each other)

\rm m' = -\frac{3}{7}

We know the:

\rm (y-y')=m'(x-x')

Where (x', y') = (7, 6), we get:

\rm (y-6)=-\frac{3}{7} (x-7)

\rm 7(y-6)=-3 (x-7)\\\\\rm 7y-42= -3x+21\\\\\rm 7y= -3x+21+42\\\\\rm 3x+7y=63

\rm -1.5x-3.5y=-31.5   (multiply by -1/2 on both sides)

Thus, the equation of the central street PQ is  -1.5x - 3.5y = -31.5

Learn more about the straight line.

brainly.com/question/3493733

6 0
1 year ago
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