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3 years ago

What number is equivalent to 3/4 b-2/3 b =-9

1 answer:

5 0

Hi,

$b = - 9 \\ \frac{3}{4} \times b - \frac{2}{3} \\ 0.75 \times - 9 - 0.67 \\ - 6.75 - 0.67 = - 7.42$

Answer: -7.42

Hope this helps.
r3t40

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3 years ago

Find the area of a rectangle whose perimeter is 60in and width is 10in.

Dafna1 [17]

Answer:

250 sq. inch

Step-by-step explanation:

Perimeter: 60 inch

: 2 X L+B

: 2 X ? + 10

Length: 60-10

: 50/2

: 25 inch

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: 25 X 10

: 250 sq.inch

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3 0

2 years ago

Which of the following expressions are equivalent to -8/11-3/4-1/4.Choose 3 answers

Effectus [21]

Answer:

option a

option d

option e

Step-by-step explanation:

option a

$-\frac{8}{11}+( -\frac{3}{4})+( -\frac{1}{4})=-\frac{8}{11}-\frac{3}{4} -\frac{1}{4}$

option d

$-(\frac{8}{11}+\frac{3}{4}+\frac{1}{4})=-\frac{8}{11}-\frac{3}{4} -\frac{1}{4}$

option e

$-\frac{8}{11}-(\frac{3}{4}+\frac{1}{4})=-\frac{8}{11}-\frac{3}{4} -\frac{1}{4}$

4 0

3 years ago

How do you solve this can someone help me pls

horrorfan [7]

Answer:

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5 0

3 years ago

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

3 years ago

Read 2 more answers