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2 years ago

$.87 per day is equal to how many dollars per year

Mathematics

2 answers:

Lynna [10]2 years ago

5 0

. 87 times 365 days =317.55 because there are 365 days in a year so 87 cents everyday for a year is. 87 times 365

Sophie [7]2 years ago

3 0

.87 × 365 = $317.55 :)

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How do I do this? Tell me how you got the answer

kotykmax [81]

1: supplementary means 180 degrees so subtract 30 from 180 and angle b is 150 degrees.
2: complementary means 90 degrees so subtract 49 from 90 and the second angle is 41 degrees.
3: a right angle is 90 degrees.
4: congruent means same size same shape so if the first angle is 40 degrees then the second angle would also be 40 degrees.

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:

Required: Solve the expressions using proper operation order

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

2 years ago

Draw a conclusion is the statement below always sometimes or never true give at least two examples to support your reasoning the

hodyreva [135]

It's sometimes true.

One example is the least common multiple of 2 and 3 is 6, which is their product.

But the product isn't always the answer because (example 2:) the least common multiple of 6 and 10 is 30 because 6*5=30 and 3*10=30, however 6*10 is 60.

Ergo, it is only sometimes true.

3 0

3 years ago

How to solve radical numerator fractions

Nonamiya [84]

Y=radical 3
just cancel out the 2 on both sides

3 0

2 years ago

Solve the inequality. The solution is....

MAVERICK [17]

Answer:

Step-by-step explanation:

-x - 6 > 21 - 28x

27x - 6 > 21

27x > 27

x > 1

8 0

3 years ago