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3 years ago

Help please and thankssssssssssssssss

Mathematics

1 answer:

maksim [4K]3 years ago

4 0

Answer:

the fourth one

trapezoid has four(4) sides

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Identify an acute angle and give its measure.

melamori03 [73]

Answer:

k&F 20 degrees

Step-by-step explanation:

hope it helps

6 0

2 years ago

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Can someone help me with this i need it for today

N76 [4]

Answer: 2y=-6.4 = y=−3.2

-4y= -3.2 = y=0.8

Step-by-step explanation:

only answering what i can sorry

3 0

3 years ago

Find the y intercept. I already found the slope is 9.

jeka57 [31]

The slope-intercept form: y = mx + b

We have the slope m = 9, therefore: y = 9x + b.

Next. We have the table:

x | -5 | -2 | 1 | 3 | 4 |

y |-46|-19| 8 |26|35|

(1, 8) → x = 1, y = 8

substitute the values of x and y to the equation:

8 = 9(1) + b

8 = 9 + b |-9

b = -1

Answer: y = 9x - 1 → y-intercept is -1.

8 0

4 years ago

<img src="https://tex.z-dn.net/?f=%20%20%5Cfrac%7B3%7D%7B4%7D%20x%20%2B%20%20%5Cfrac%7B5%7D%7B6%7D%20%20%3D%205x%20-%20%20%5Cfra

jonny [76]

Answer:

nos agoi o se

Step-by-step explanation:

6 0

3 years ago

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Which is the simplified form of the expression ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript nega

AlekseyPX

Answer:

The option "StartFraction 1 Over 3 Superscript 8" is correct

That is $\frac{1}{3^8}$ is correct answer

Therefore $[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}$

Step-by-step explanation:

Given expression is ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript negative 3 Baseline times ((2 Superscript negative 3 Baseline) (3 squared)) squared

The given expression can be written as

$[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2$

To find the simplified form of the given expression :

$[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2$

$=(2^{-2})^{-3}(3^4)^{-3}\times (2^{-3})^2(3^2)^2$ ( using the property $(ab)^m=a^m.b^m$ )

$=(2^6)(3^{-12})\times (2^{-6})(3^4)$ ( using the property $(a^m)^n=a^{mn}$

$=(2^6)(2^{-6})(3^{-12})(3^4)$ ( combining the like powers )

$=2^{6-6}3^{-12+4}$ ( using the property $a^m.a^n=a^{m+n}$ )

$=2^03^{-8}$

$=\frac{1}{3^8}$ ( using the property $a^{-m}=\frac{1}{a^m}$ )

Therefore $[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}$

Therefore option "StartFraction 1 Over 3 Superscript 8" is correct

That is $\frac{1}{3^8}$ is correct answer

6 0

4 years ago

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