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3 years ago

12

Help please my brain isn’t working today I promise I would normally know this HELP! Worth 40 points:)

2 answers:

zubka84 [21]3 years ago

6 0

Answer:

2/15

...................

satela [25.4K]3 years ago

3 0

Answer:2/15

Step-by-step explanation:

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Find value of variable

Nezavi [6.7K]

<span>Step 1: Identify the variables.
Step 2: Add/Subtract whole numbers so they're all on one side.
Step 3: Add/Subtract Variables so they're all on one side.
Step 4: Multiply /Divide to get the variable by itself.
<span>Step 5: Check your work.</span></span>

4 0

4 years ago

Use a special product pattern to evaluate <img src="https://tex.z-dn.net/?f=19%5E2-11%5E2" id="TexFormula1" title="19^2-11^2" al

Delvig [45]

Answer:

240

Step-by-step explanation:

To evaluate 19^2 - 11^2, we have to use a special product pattern. Since this expression is in the form a^2 - b^2, we use the pattern (a + b)(a - b).

$(19+11)(19-11)$
$(30)(8)$
$240$

Therefore, the answer is 240.

Have a lovely rest of your day/night, and good luck with your assignments! ♡

3 0

3 years ago

3x+5-x=2x+7 is infinite real soluction, no real soluction or one real soluction

Helga [31]

Answer:

There is no real solution.

Step-by-step explanation:

There is no solution because 5 is not equal to 7.

8 0

3 years ago

Find the break-even point for the given cost and revenue equations. Round to the nearest whole unit. C = 94n + 534,000 R = 168n

elena-14-01-66 [18.8K]

Answer:

The will be answer is 7216

Step-by-step explanation:

3 0

3 years ago

For the figures below, assume they are made of semicircles, quarter circles and squares. For each shape, find the area and perim

Marina86 [1]

Area of the shaded region $$=36(\pi -2)$ square cm

Perimeter of the shaded region $=6 (\pi + 2\sqrt 2)$ cm

Solution:

Radius of the quarter of circle = 12 cm

Area of the shaded region = Area of quarter of circle – Area of the triangle

$$=\frac{1}{4} \pi r^2 - \frac{1}{2} bh$

$$=\frac{1}{4} \pi \times 12^2 - \frac{1}{2} \times 12 \times 12$

$$=36\pi -72$

$$=36(\pi -2)$ square cm.

Area of the shaded region $$=36(\pi -2)$ square cm

Using Pythagoras theorem,

$AC^2=AB^2+BC^2$

$AC^2=12^2+12^2$

$AC^2=288$

Taking square root on both sides of the equation, we get

$AC= 12\sqrt 2$ cm

Perimeter of the quadrant of a circle = $\frac{1}{4} \times 2\pi r$

$$=\frac{1}{4} \times 2 \times \pi \times 12$

$$=6 \pi$ cm

Perimeter of the shaded region = $6 \pi + 12\sqrt 2$ cm

$=6 (\pi + 2\sqrt 2)$ cm

Hence area of the shaded region $$=36(\pi -2)$ square cm

Perimeter of the shaded region $=6 (\pi + 2\sqrt 2)$ cm

6 0

3 years ago