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3 years ago

Anyone know the answer?

Mathematics

1 answer:

balu736 [363]3 years ago

5 0

Answer:

segment BC is located at B (1, 0) and C (1, 3) and is one-half the size of segment B C .

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Johnson High School has a total enrollment of 1829 students. Season tickets are sold for the basketball season and 604 students

Nezavi [6.7K]

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The formula is number of students who bought the tickets
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.......................number of students enrolled

8 0

3 years ago

The points (2, 3) and (6, 11) lie on a line. What is the slope of this line?

AveGali [126]

Answer:

8/4

Step-by-step explanation:

to do slope is y2 - y1 over x2 - x1, so it would be 11 - 3 for 8, then 6 - 2, leaving 8/4.

3 0

3 years ago

Divide.<br> Write your answer in simplest form.<br> 6/5 divided by 5/8

Anna007 [38]

6/5x8/5

6x8 5x5

48/5x5

48/25

Decimal (1.92)

5 0

3 years ago

Read 2 more answers

Jacob has 1/2 of a container of trail mix. He is filling snack packs that each use about 1/6 of a container. How many snack pack

kondaur [170]

Answer:

x = 3

Step-by-step explanation:

Jacob has 1/2 of a container of trail mix.

He is filling snack packs that each use about 1/6 of a container

We need to find the number of snack packs can Jacob make. Let he can make x snack packs.

$\dfrac{1}{6}x=\dfrac{1}{2}\\\\x=3$

Hence, he can make 3 snack packs.

3 0

3 years ago

Simplify each expression using the proper order of operations. please help thank you so much :)

Tasya [4]

Answer:

$\frac{41 - 3^2}{\sqrt{36} * 3 - 26} = -4$

$12 + 3\sqrt{8} * (9 - 2) = 12 + 42 \sqrt{2}$

$\frac{28 - (7^2 + 3)}{-13 + 3 * 5} = -12$

$7^2 - 5* 8+1 = 10$

$(2 * \sqrt{16}) - (\sqrt[3]{27} * \sqrt{81}) + 1 = -18$

Step-by-step explanation:

<em>Required: Solve the expressions using proper operation order</em>

To solve this, we'll make use of BODMAS

--------------------------------------------------------------------------------------------------------

$\frac{41 - 3^2}{\sqrt{36} * 3 - 26}$

Evaluate all squares and square roots

$\frac{41 - 3*3}{\sqrt{36} * 3 - 26}$

$\frac{41 - 3*3}{6 * 3 - 26}$

Evaluate the numerator (Start by multiplying 3 * 3)

$\frac{41 - 9}{6 * 3 - 26}$

Subtract 9 from 41

$\frac{32}{6 * 3 - 26}$

Evaluate the denominator (Start by multiplying 6 * 3)

$\frac{32}{18 - 26}$

$\frac{32}{-8}$

Divide 32 by -8

-4

Hence;

$\frac{41 - 3^2}{\sqrt{36} * 3 - 26} = -4$

--------------------------------------------------------------------------------------------------------

$12 + 3\sqrt{8} * (9 - 2)$

Start by evaluating the bracket

$12 + 3\sqrt{8} * 7$

Then evaluate the multiplication

$12 + 21\sqrt{8}$

Simplify the square root

$12 + 21\sqrt{4 * 2}$

Split the square root

$12 + 21\sqrt{4} * \sqrt{2}$

Take Square root of 4

$12 + 21 * 2} * \sqrt{2}$

$12 + 42 \sqrt{2}$

Hence;

$12 + 3\sqrt{8} * (9 - 2) = 12 + 42 \sqrt{2}$

--------------------------------------------------------------------------------------------------------

$\frac{28 - (7^2 + 3)}{-13 + 3 * 5}$

Evaluate 7²

$\frac{28 - (49 + 3)}{-13 + 3 * 5}$

Evaluate all expression in the bracket

$\frac{28 - (52)}{-13 + 3 * 5}$

$\frac{28 - 52}{-13 + 3 * 5}$

Evaluate 3 * 5

$\frac{28 - 52}{-13 + 1 5}$

$\frac{-2 4}{2}$

$-12$

Hence;

$\frac{28 - (7^2 + 3)}{-13 + 3 * 5} = -12$

--------------------------------------------------------------------------------------------------------

$7^2 - 5* 8+1$

Evaluate 7²

$49 - 5* 8+1$

Evaluate 5 * 8

$49 - 40 + 1$

$10$

$7^2 - 5* 8+1 = 10$

--------------------------------------------------------------------------------------------------------

$(2 * \sqrt{16}) - (\sqrt[3]{27} * \sqrt{81}) + 1$

Evaluate all square root and cube root

$(2 * 4) - (3 * 9) + 1$

Solve the expressions in bracket

$8 - 27 + 1$

$-18$

Hence;

$(2 * \sqrt{16}) - (\sqrt[3]{27} * \sqrt{81}) + 1 = -18$

7 0

3 years ago