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3 years ago

Please need help evaluate the math problems

Mathematics

1 answer:

alina1380 [7]3 years ago

6 0

x + y when x = 6 and y = 2

6 + 2 = 8

$\frac{9x+24}{4}$ when x = 12 and y = 20

$\frac{148}{4}$ = 37

$\frac{6m}{n}$ when m = 12 and n = 3

$\frac{72}{3}$ = 24

Six times x

Six increased by x

6 + x

The sum of x and y

x + y

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How do you write 5.9 in expanded form

tresset_1 [31]

Answer:
(5 * 1) + (9 * 0.1).

Explanation:
To write a number in expanded form you have each digit multiplied by its numbers place and add each digit together. 5 is in the ones place so it becomes (5 * 1).
9 is in the tenths place so it becomes (9 * 0.1).
add them together to get (5 * 1) + (9 * 0.1)

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3 years ago

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Simplify 4.51^0 - which is it? 4.51 ? 1 ? 0 ? -1 ?

koban [17]

Answer:

$1$

Step-by-step explanation:

${(4.51) }^{0} = 1$

hope this helps you.

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3 years ago

Any genius out there to solve this?

Helga [31]

Answer:

Step-by-step explanation:

The pattern is + 1, + 3, + 6,... i.e. + 1, + (1 + 2), + (1 + 2 + 3),...

.'. Missing number = 15 + (1 + 2 + 3 + 4) = 25.

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3 years ago

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Simplify -7(-4p+1) + 2p

mariarad [96]

Step-by-step explanation:

-7(-4p+1) + 2p

-7 x -4 = 28p

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28p + 2p = 30p

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3 years ago

Fill in Sin, Cos, and tan ratio for angle x. Sin X = 4/5 (28/35 simplified)

Fantom [35]

Answer:

Given: $\sin(x) = (4/5)$ .

Assuming that $0 < x < 90^{\circ}$ , $\cos(x) = (3/5)$ while $\tan(x) = (4/3)$ .

Step-by-step explanation:

By the Pythagorean identity $\sin^{2}(x) + \cos^{2}(x) = 1$ .

Assuming that $0 < x < 90^{\circ}$ , $0 < \cos(x) < 1$ .

Rearrange the Pythagorean identity to find an expression for $\cos(x)$ .

$\cos^{2}(x) = 1 - \sin^{2}(x)$ .

Given that $0 < \cos(x) < 1$ :

$\begin{aligned} &\cos(x) \\ &= \sqrt{1 - \sin^{2}(x)} \\ &= \sqrt{1 - \left(\frac{4}{5}\right)^{2}} \\ &= \sqrt{1 - \frac{16}{25}} \\ &= \frac{3}{5}\end{aligned}$ .

Hence, $\tan(x)$ would be:

$\begin{aligned}& \tan(x) \\ &= \frac{\sin(x)}{\cos(x)} \\ &= \frac{(4/5)}{(3/5)} \\ &= \frac{4}{3}\end{aligned}$ .

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2 years ago