The quotient of the number given number 7 Superscript negative 1 Baseline Over 7 Superscript negative 2 Baseline is 7.
<h3>What is the quotient?</h3>
Quotient is the resultant number which is obtain by dividing a number with another. Let a number a is divided by number b. Then the quotient of these two number will be,
Here, (<em>a, b</em>) are the real numbers.
The number StartFraction 7 Superscript negative 1 Baseline Over 7 Superscript negative 2 Baseline EndFraction, given can be written as,
Let the quotient of this division is n. Therefore,
A number in numerator of a fraction with negative exponent can be written in the denominator with the same but positive exponent and vise versa. Therefore,
Hence, the quotient of the number given number 7 Superscript negative 1 Baseline Over 7 Superscript negative 2 Baseline is 7.
Learn more about the quotient here;
brainly.com/question/673545
Answer:
t = 6 s
Step-by-step explanation:
Given that,
Sherry is driving 390 miles to visit The gateway arch in St. Louis.
She drives at an average rate of 65 miles per hour.
We need to find the amount of time it will take Sherry to get to the arch. Let the time is t.
Speed = distance/time
Hence, it will take 6 hours to get to the arch.
Answer:
11
12
does not
Step-by-step explanation:
If we know how much money the coach has and how much each bag costs we can find out how much bags the coach can by by divide the money the coach has by the price of one bat bag. So we do....
605 / 55 = 11
So now we know that she can by 11 bags maximum, while she needs 12 bags for the team. From this we know that she does not have enough money to buy 12 bags.
Answer:
x-y=4
and
x+y=32 is <u>your</u><u> </u><u>equation</u>
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<u>s</u><u>u</u><u>m</u><u> </u><u>o</u><u>f</u><u> </u><u>t</u><u>w</u><u>o</u><u> </u><u>n</u><u>u</u><u>m</u><u>b</u><u>e</u><u>r</u><u>=</u><u>x</u><u>+</u><u>y</u><u> </u><u>i</u><u>s</u><u> </u><u>e</u><u>q</u><u>u</u><u>a</u><u>l</u><u> </u><u>t</u><u>o</u><u> </u><u>3</u><u>2</u>
<u>x</u><u>-</u><u>y</u><u>=</u><u>4</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>(</u><u>1</u><u>)</u>
<u>x</u><u>+</u><u>y</u><u>=</u><u>3</u><u>2</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>(</u><u>2</u><u>)</u>
<u>a</u><u>d</u><u>d</u><u>i</u><u>n</u><u>g</u><u> </u><u>equation</u><u> </u><u>1</u><u> </u><u>a</u><u>n</u><u>d</u><u> </u><u>2</u>
<u>x</u><u>-</u><u>y</u><u> </u><u>+</u><u>x</u><u>+</u><u>y</u><u>=</u><u>4</u><u>+</u><u>3</u><u>2</u>
<u>2</u><u>x</u><u>=</u><u>3</u><u>6</u>
<u>x</u><u>=</u><u>3</u><u>6</u><u>/</u><u>2</u><u>=</u><u>1</u><u>6</u><u>a</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u>
<u>s</u><u>u</u><u>b</u><u>s</u><u>t</u><u>i</u><u>t</u><u>u</u><u>t</u><u>i</u><u>n</u><u>g</u><u> </u><u>value</u><u> </u><u>of</u><u> </u><u>y</u><u> </u><u>in</u><u> </u><u>equation</u><u> </u><u>1</u>
<u>1</u><u>6</u><u>-</u><u>y</u><u>=</u><u>4</u>
<u>1</u><u>6</u><u>-</u><u>4</u><u>=</u><u>y</u>
<u>y</u><u>=</u><u>1</u><u>2</u><u> </u><u>a</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u>
