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2 years ago

12

Kathy measured her finger using a ruler with centimeters then she measured the same finger with a ruler using millimeters can Ka

thy compare the two measurements

1 answer:

mihalych1998 [28]2 years ago

7 0

Answer:

Yes, 10 millimeters is the same as 1 centimeter.

Step-by-step explanation:

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Over the last 3 months, Janet exercised a total of 101 hours. She exercised 4 hours more during the second month than the first

olganol [36]

In 1st month let she exercised h hours

in second month she exercised h+4 hours

in third month she exercised h+4-3=h+1 hours

Now

h+h+4+h+1=101
3h+5=101
3h=96
h=32hours

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2 years ago

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Your Co. collects 50% of its sales in the month of the sales, 30% of the follow month, and 20% the second month after the sale.

Crank

Answer:

March 52
April 41

Step-by-step explanation:

In March, Your Co. will collect 20% of January's sales, 30% of February's sales, and 50% of March's sales:

.20×50 +.30×40 +.50×60 = 10 +12 +30 = 52

Similarly, in April, collections will be ...

.20×40 + .30×60 + .50×30 = 8 +18 +15 = 41

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3 years ago

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Mia recorded the hair colors of the people who have already taken their seats at a seminar.

Aloiza [94]

Answer:

14

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1 year ago

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Someone answer. Will give brainlest

Leto [7]

Area inside the semi-circle and outside the triangle is (91.125π - 120) in²

Solution:

Base of the triangle = 10 in

Height of the triangle = 24 in

Area of the triangle = $\frac{1}{2} bh$

$$=\frac{1}{2} \times 10\times24$

Area of the triangle = 120 in²

Using Pythagoras theorem,

$\text{Hypotenuse}^2=\text{base}^2+\text{height}^2$

$\text{Hypotenuse}^2=10^2+24^2$

$\text{Hypotenuse}^2=100+576$

$\text{Hypotenuse}^2=676$

Taking square root on both sides, we get

Hypotenuse = 23 inch = diameter

Radius = 23 ÷ 2 = 11.5 in

Area of the semi-circle = $\frac{1}{2}\pi r^2$

$$=\frac{1}{2} \pi \times (13.5)^2$

Area of the semi-circle = 91.125π in²

Area of the shaded portion = (91.125π - 120) in²

Area inside the semi-circle and outside the triangle is (91.125π - 120) in².

5 0

3 years ago

Solve the equation: <img src="https://tex.z-dn.net/?f=5%5Ex%5E%28%5Ex%5E-%5E1%5E%29%3D1" id="TexFormula1" title="5^x^(^x^-^1^)=1

Tomtit [17]

<u>The concept:</u>

We are given the equation:

$5^{x(x-1)} = 1$

Which can be simplified as:

$5^{x^{2} - x} = 1^{1}$

Since any number to the power '0' is 1

x² - x must be equal to 0 for the given equation to be true

<u></u>

<u>Solving for x:</u>

x² - x = 0

x(x-1) = 0

now, we can divide both sides by either x OR x-1

So we will see what we get for either choice:

x = 0/(x-1) x-1 = 0/x

x = 0 x = 1

Hence, the value of x is either 0 or 1

7 0

3 years ago