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3 years ago

What are two different ways for finding the area of the shaded region. All angles are right angles.

Mathematics

2 answers:

Vika [28.1K]3 years ago

3 0

Answer:

one way is L times W (length times width)

Step-by-step explanation:

for example if you had 5ft. for length and 6ft. for width you would times the two to get you area

Zinaida [17]3 years ago

3 0

L times w...............

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HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Radda [10]

This is what the diagram is saying
(-4)+(-9)= -13

6 0

2 years ago

If <img src="https://tex.z-dn.net/?f=36%20%5E%7B12%20-%20m%7D%20%3D%206%5E%7B2m%7D%20" id="TexFormula1" title="36 ^{12 - m}

Katyanochek1 [597]

Answer:

Step-by-step explanation:

$6^{2(12-m)}=6^{2m}\\2(12-m)=2m\\24-2m=2m\\24=4m\\m = 6$

8 0

3 years ago

20 people applied for a job. Everyone either has a school certificate or diploma or even both. If 14 have school certificates an

STatiana [176]

Given:

Either has a school certificate or diploma or even both = 20 people

Having school certificates = 14

Having diplomas = 11

To find:

The number of people who have a school certificate only.

Solution:

Let A be the set of people who have school certificates and B be the set of people who have diplomas.

According to the given information, we have

$n(A)=14$

$n(B)=11$

$n(A\cup B)=20$

We know that,

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$

$20=14+11-n(A\cap B)$

$20=25-n(A\cap B)$

Subtract both sides by 25.

$20-25=-n(A\cap B)$

$-5=-n(A\cap B)$

$5=n(A\cap B)$

We need to find the number of people who have a school certificate only, i.e. $n(A\cap B')$ .

$n(A\cap B')=n(A)-n(A\cap B)$

$n(A\cap B')=14-5$

$n(A\cap B')=9$

Therefore, 9 people have a school certificate only.

3 0

3 years ago

A metalworker has a metal alloy that is 15% copper and another alloy that is 75% copper. How many kilograms of each alloy shou

lisov135 [29]

48 kilograms of 15 % copper is combined with 72 kilograms of 75 % copper to create 120 kg of a 51% copper alloy

Solution:

Let "x" be the kilograms of 15 % copper

Then, (120 - x) be the kilograms of 75 % copper

Then, according to question,

"x" kilograms of 15 % copper is combined with (120 - x) kilograms of 75 % copper to create 120 kg of a 51% copper alloy

Thus we frame a equation as:

15 % of x + 75 % of (120 - x) = 51 % of 120

Solve the equation for "x"

$15 \% \times x + 75 \% \times (120-x) = 51 \% \times 120\\\\\frac{15}{100} \times x + \frac{75}{100} \times (120-x) = \frac{51}{100} \times 120\\\\0.15x + 0.75(120-x) = 0.51 \times 120\\\\0.15x + 90 - 0.75x = 61.2\\\\0.6x = 90 - 61.2\\\\0.6x = 28.8\\\\Divide\ both\ sides\ by\ 0.6\\\\x = 48$

Thus 48 kilograms of 15 % copper used

Then, (120 - x) = (120 - 48) = 72

Thus, 72 kilograms of 75 % copper is used

4 0

3 years ago

An airplane has begun its descent for a landing. When the airplane is 150 miles west of its destination, its altitude is 25,000

belka [17]

The airplane has descended (25,000 - 19,000) = 6,000 feet
while flying (150 - 90) = 60 miles.

If the descent is modeled by a linear function, then the slope
of the function is

(-6000 ft) / (60 miles) = - 100 ft/mile .

Since it still has 19,000 ft left to descend, at the rate of 100 ft/mi,
it still needs to fly

(19,000 ft) / (100 ft/mile) = 190 miles

to reach the ground.

It's located 90 miles west of the runway now. So if it continues
on the same slope, it'll be 100 miles past the runway (east of it)
when it touches down.

I sure hope there's another airport there.

5 0

3 years ago

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