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3 years ago

Maths hw???????please help^^^^^^^^

Mathematics

1 answer:

belka [17]3 years ago

7 0

6(2x+1) ≤ 5x-1
Distribute
12x+6 ≤ 5x-1
-5x -5x

7x+6 ≤ -1
-6 -6

7x ≤ -7
/7 /7

x ≤ -1

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Calculate the new balance using the previous balance method. Previous balance = $34.80 Finance charge = $0.75 New purchases = $8

pychu [463]

<h3>Given</h3>

new balance = previous balance + finance charge + purchases - payments

previous balance = $34.80

finance charge = $0.75

purchases = $83.21

payments = $5.50

new balance

<h3>Solution</h3>

Fill in the given information and do the arithmetic.

... new balance = previous balance + finance charge + purchases - payments

... new balance = $34.80 + 0.75 + 83.21 - 5.50

... new balance = $113.26 . . . . matches the 2nd choice

7 0

3 years ago

if y represents total earnings in dollars and x represents hours worked, then which equation models the wages of someone who mak

Klio2033 [76]

Answer: $y=11.50x$

Step-by-step explanation:

1. You have the following information given in the problem above:

- y represents total earnings in dollars .

- x represents hours worked.

2. Keeping on mind that the expression must model the wages of someone who makes $11.50 in an hour, this amount of money per hour must multiply the hours worked, which is represented by x.

3. Therefore, you can write the following equation:

$y=11.50x$

5 0

3 years ago

A person stands 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 meters

In-s [12.5K]

Answer:

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

Step-by-step explanation:

Given that,

A person stand 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 m/s.

From Pythagorean Theorem,

(The distance between car and person)²= (The distance of the car from intersection)²+ (The distance of the person from intersection)²+

Assume that the distance of the car from the intersection and from the person be x and y at any time t respectively.

∴y²= x²+10²

$\Rightarrow y=\sqrt{x^2+100}$

Differentiating with respect to t

$\frac{dy}{dt}=\frac{1}{2\sqrt{x^2+100}}. 2x\frac{dx}{dt}$

$\Rightarrow \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}. \frac{dx}{dt}$

Since the car driving towards the intersection at 13 m/s.

so, $\frac{dx}{dt}=-13$

$\therefore \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}.(-13)$

Now

$\therefore \frac{dy}{dt}|_{x=24}=\frac{24}{\sqrt{24^2+100}}.(-13)$

$=\frac{24\times (-13)}{\sqrt{676}}$

$=\frac{24\times (-13)}{26}$

= -12 m/s

Negative sign denotes the distance between the car and the person decrease.

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

8 0

3 years ago

How do I factor . X2 { squared } -9x+28?

Dmitrij [34]

It is not factorable

3 0

3 years ago

I GIVE BRAINLIEST PLS HELP

ioda

The answer is 8
Here's why:
${ ( \frac{( {6}^{7}) \times ( {3}^{3}) }{( {6}^{6}) \times ( {3}^{4} ) } )}^{3} = \\ ( \frac{6}{3} ) ^{3} = \\ \frac{216}{27} = 8$
The exponents are subtracted one from another when divided.
$\frac{ a ^{b} }{ {a}^{c} } = {a}^{b - c}$
We can look at the problem this way:
$( \frac{6^{7} }{6 ^{6} } \times \frac{3^{3} }{ {3}^{4} } ) = (6^{7 - 6} \times {3}^{3 - 4} ) = \\ ({6}^{1} \times {3}^{ - 1} )$
Since we have the power of -1 on the 3, we apply this rule:
${a}^{ - b} = \frac{1}{ {a}^{b} }$
Also this rule because we have the power of 1 on the 6:
${a}^{1} = a$
Then we get this:
$(6 \times \frac{1}{3} )^{3} = ( \frac{6}{3} )^{3}$
We apply the rule:
$( \frac{a}{b}) ^{c} = \frac{ {a}^{c} }{ {b}^{c} }$
We get this:
$\frac{{6}^{3} }{ {3}^{3} } = \frac{216}{27} = 8$

4 0

3 years ago