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Kazeer [188]
3 years ago
14

The function g is defined by g(x)=9k−4, where k is a constant. Find k, if the graph of g passes through the point (7,−2).

Mathematics
2 answers:
Tomtit [17]3 years ago
7 0
<span>given g(x)=9k−4 and </span><span>g passes through the point (7,−2)

substiting g(7) = 9*k - 4 = -2

so 9*k = -2 + 4 = 2

k = 2/9 or 0.22</span>
Nutka1998 [239]3 years ago
3 0

Answer: k=\dfrac{2}{9}

Step-by-step explanation:

Given : The function g is defined by g(x)=9k-4, where k is a constant.

If the graph of g passes through the point (7,−2), then at x=7 , the value of g should be -2.

So substitute x=7 then g(7)=-2.

Consider g(7)=9k-4

\Rightarrow\ -2=9k-4

Adding 4 on both sides , we get

\Rightarrow\ 9k=2

Dividing both sides  by 9 , we get

k=\dfrac{2}{9}

Hence, the value of k=\dfrac{2}{9}.

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This was based on my work I did.

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Solve and verify please solve​
RUDIKE [14]

Answer:

Below in bold.

Step-by-step explanation:

I'll solve a few of them to illustrate the method:

a) 5x - (3x - 1) = x - 4

The aim is to get all the 'x' terms to one sides of the '=' and the numbers to the other side:

Distribute the negative over the 3x - 1 in the parentheses:

5x - 3x + 1 = x - 4        ( Note that  -* -1 = + 1)

Now subtract  1 from both sides:

5x - 3x + 1 - 1 = x - 4 - 1

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Now subtract x from both sides:

5x - 3x - x = x - x - 5.

2x - x = -5

x = -5.

(b) 3/5 x + 1 = 2/5 - 3x

We first get rid of the fractions by multiplying each term by 5:

5 * 3/5 x + 1*5 = 5 * 2/5 - 5*3x

3x + 5 = 2 - 15x

3x + 15x + 5 = 2 - 15x + 15x        (adding 15x to both sides)

18x + 5 = 2          Subtract 5 from both sides:

18x + 5 - 5 = 2 - 5

18x = -3

Now we divide both sides by 18:

x = -3/18 = -1/6.

g) 0.3(6 + m) = 0.4(8 - m)

0.3 * 6 + 0.3* m = 0.4*8 + 0.4 * -m   ( Distributing the 0.3 and 0.4 over the parentheses)

1.8 + 0.3m = 3.2 - 0.4m        ( recall that + * - = -)

1.8 - 1.8 + 0.3m = 3.2 - 1.8 - 0.4m

0.3m = 1.4 - 0.4m

0.3m + 0.4m = 1.4 - 0.4m + 0.4m

0.7m = 1.4

m = 1/4/0.7

m =  2.

I hope this helps.

5 0
3 years ago
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