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2 years ago

WILL GIVE BRAINLIEST HELP URGENT!!!

Mathematics

1 answer:

denis23 [38]2 years ago

8 0

Answer:

Find like terms:

5x and 2x are like terms,

as well as:

42 and 1

do the math per usual but with the like terms:

5x - 2x = 3x

42 -1 = 41

Therefore your answer is:

(3x - 41)

(hope this helps)

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1/2 log2x=3log23 rewrite with out using logarithmic

Brums [2.3K]

Answer:

729

Step-by-step explanation:

$\frac{1}{2}log_{2}x=3log_{2}3 \\$

$log_{2}\sqrt{x} =log_{2}3^{3}$

$\sqrt{x} =27 \\$

x=27²=729

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2 years ago

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Stefan spent half of his weekly allowance on clothes to earn more money his parents let him wash the car for $5 what is his week

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2 years ago

Student work<br> TRY IT #1<br> Write each of the following as a rational number.<br> b. 3<br> c-4

tino4ka555 [31]

Answer:

b. 3 = 3/1

c. -4 = -4/1

Step-by-step explanation:

An integer can be written as a rational number with a denominator of 1. The denominator can be anything else, too, in which case the numerator must be multiplied by that same value.

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3 years ago

The total mass of the Sun is about 2×10^30 kg, of which about 76 % was hydrogen when the Sun formed. However, only about 12 % of

nignag [31]

Answer:

A. 1.8 × $10^{30}$ Kg

B i. 3.0 × $10^{17}$ seconds

ii. 9.6 × $10^{9}$ years

C. After 9.2 × $10^{9}$ (9.2 billion) years

Step-by-step explanation:

Given that the mass of the Sun = 2× $10^{30}$ Kg.

Mass of hydrogen when Sun was formed = 76% of 2× $10^{30}$ Kg

= $\frac{76}{100}$ ×2× $10^{30}$ Kg

= 1.52 × $10^{30}$ Kg

Mass of hydrogen available for fusion = 12% of 1.52 × $10^{30}$ Kg

= $\frac{12}{100}$ × 1.52 × $10^{30}$ Kg

= 1.824 × $10^{30}$ Kg

A. Total mass of hydrogen available for fusion over the lifetime of the sun is 1.8 × $10^{30}$ Kg.

B. Given that the Sun fuses 6 × $10^{11}$ Kg of hydrogen each second.

i. The Sun's initial hydrogen would last;

$\frac{1.8*10^{30} }{6*10^{11} }$

= 3.04 × $10^{17}$ seconds

The Sun's hydrogen would last 3.0 × $10^{17}$ seconds

ii. Since there are 31536000 seconds in a year, then;

The Sun's initial hydrogen would last;

$\frac{3.04*10^{17} }{31536000}$

= 9.640 × $10^{9}$ years

The Sun's hydrogen would last 9.6 × $10^{9}$ years.

C. Given that our solar system is now about 4.6 × $10^{9}$ years, then;

$\frac{9.6*10^{9} }{4.6*10^{9} }$

= 2.09

So that; 2 × 4.6 × $10^{9}$ = 9.2 × $10^{9}$ years

Therefore, we need to worry about the Sun running out of hydrogen for fusion after 9.2 × $10^{9}$ years.

6 0

3 years ago