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aev [14]
3 years ago
14

What is the product of 5.86 × 10–7 and 3.1 × 104 What is the value of n in the scientific notation of the product?

Mathematics
2 answers:
il63 [147K]3 years ago
7 0

Answer:

-2

Step-by-step explanation:

2020 prrrrrrrrrr

sveticcg [70]3 years ago
6 0

the answer is -2 on edge your welcome

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Does anyone know how to do this please help me
Anit [1.1K]

Answer:

162°

Step-by-step explanation:

a circle has 360°, or the angle of a whole circle is that

so 45% of 360 =

0.45 times 360 = 162

7 0
3 years ago
ASAP PLEASE HELP! THANK YOUU​
mote1985 [20]
The answer to the question is c
4 0
3 years ago
-4rover4+5rover3 please help me
Alchen [17]

Solution, \frac{-4r}{4}+\frac{5r}{3}=\frac{2r}{3}

Steps:

\frac{-4r}{4}+\frac{5r}{3}

\frac{-4r}{4},\\\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b},\\-\frac{4r}{4},\\\mathrm{Divide\:the\:numbers:}\:\frac{4}{4}=1,\\-r,\\-r+\frac{5r}{3}

\mathrm{Convert\:element\:to\:fraction}:\quad \:r=\frac{r3}{3},\\\frac{5r}{3}-\frac{r\cdot \:3}{3}

\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c},\\\frac{5r-r\cdot \:3}{3}

5r-r\cdot \:3,\\\mathrm{Add\:similar\:elements:}\:5r-3r=2r,\\\frac{2r}{3}

The\:Correct\:Answer\:Is\:\frac{2r}{3}

Hope\:This\:Helps!!!

-Austint1414

5 0
3 years ago
In ΔJKL, \overline{JL} JL is extended through point L to point M, m∠JKL = (2x+14)^{\circ}(2x+14) ∘ , m∠LJK = (3x+15)^{\circ}(3x+
Sergio039 [100]

Answer:

The value of x =10°

Step-by-step explanation:

Check attachment for solution.

Triangle theorem: the sum of the opposite interior angle is equal to the exterior angle.

<JKL+<LJK=<KLM

2x+14+3x+15=7x+9

Collect like terms

2x+3x-7x=9-14-15

-2x=-20

Divide both side by -2

x=10°

6 0
3 years ago
Prove that root 7 is irrational by the method of contradiction
Alchen [17]

Let assume that \sqrt7 is a rational number. Therefore it can be expressed as a fraction \dfrac{a}{b} wherea,b\in\mathbb{Z} and \text{gcd}(a,b)=1.

\sqrt7=\dfrac{a}{b}\\\\7=\dfrac{a^2}{b^2}\\\\a^2=7b^2

This means that a^2 is divisible by 7, and therefore also a is divisible by 7.

So, a=7k where k\in\mathbb{Z}

(7k)^2=7b^2\\\\49k^2=7b^2\\\\7k^2=b^2

Analogically to a^2=7b^2 ------- b^2 is divisible by 7 and therefore so is b.

But if both numbers a and b are divisible by 7, then \text{gcd}(a,b)=7 which contradicts our earlier assumption that \text{gcd}(a,b)=1.

Therefore \sqrt7 is an irrational number.

8 0
3 years ago
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