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3 years ago

Solve for b. -t = 9(t - 10) (WILL GIVE BRAINLIEST)

Mathematics

2 answers:

coldgirl [10]3 years ago

5 0

Answer:

Do you mean solve for t?

If you do, the answer is 9.

Step-by-step explanation:

If we expand the equation out, we get:

-t = 9t - 90

If we move the -90 over to the left side and the -t to the right side, we get:

90 = 10t

Dividing both sides by 9, we get:

t = 9

zavuch27 [327]3 years ago

3 0

T=9
The anser is this and its easy first multiply

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Which ordered pair makes both inequalities true?

kvasek [131]

Answer:

(3, 0)

Step-by-step explanation:

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3 years ago

Two decimals numbers, to the thousandths, that have a sum of about 5 and a difference of about 3. Then do the math to prove it w

mina [271]

Answer:

The numbers are approximately: 4.000 and 1.000

Step-by-step explanation:

The given parameters can be expressed as:

$\frac{x}{1000} + \frac{y}{1000} \approx 5$

$\frac{x}{1000} - \frac{y}{1000} \approx 3$

Required

Determine the numbers

In each of the equations, multiply by 1000

$1000 * [\frac{x}{1000} + \frac{y}{1000} \approx 5]$

$x + y \approx 5000$

$1000 * [\frac{x}{1000} - \frac{y}{1000} \approx 3]$

$x-y \approx 3000$

So, we have:

$x + y \approx 5000$

$x-y \approx 3000$

Add the two equations

$x + x + y - y \approx 5000 + 3000$

$2x \approx 8000$

Solve for x

$x \approx 8000/2$

$x \approx 4000$

Substitute $x \approx 4000$ in $x + y \approx 5000$

$4000 + y \approx 5000$

$y \approx 5000 - 4000$

$y \approx 1000$

Using:

$\frac{x}{1000} + \frac{y}{1000} \approx 5$

We have:

$\frac{4000}{1000} + \frac{1000}{1000} \approx 5$

$4.000 + 1.000 \approx5$

<em>This implies that, the numbers approximates to 4.000 and 1.000, respectively.</em>

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3 years ago