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1 year ago

PLEASE LOOK AT PHOTO! WHOEVER ANSWERS CORRECTLY I WILL MARK BRAINIEST!!

Mathematics

2 answers:

Trava [24]1 year ago

4 0

Answer:

Step-by-step explanation:

Let m2 be x

Then m1 will be= 4x

A/Q

--> m1+m2= 115

4x+x= 115

5x= 115

x=115/5

x= 23

--> m2= 23 degree

--> m1= 4x

= 4*23

= 92 degree

8_murik_8 [283]1 year ago

3 0

Step-by-step explanation:

let m2 be x

then m1 = 4x

Since, m1+m2 = 115

therefore x + 4x = 115

5x = 115

x = 23

so , m2 = 23°

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Round 5,382,619 to the nearest 1000

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2 years ago

Read 2 more answers

The population of the world was 7.1 billion in 2013, and the observed relative growth rate was 1.1% per year. A) Estimate how lo

VLD [36.1K]

Answer:

A) 63.36 years.

B) 100.42 years.

Step-by-step explanation:

We have been given that the population of the world was 7.1 billion in 2013, and the observed relative growth rate was 1.1% per year.

A) Since we know that population increases exponentially, therefore we will use our given information to form an exponential model for population increase and then we will solve for the time by which our population will be double.

$P(t)=7.1(1.011)^{t}$

$7.1 \times 2=7.1(1.011)^{t}$

$\frac{7.1 \times 2}{7.1} =(1.011)^{t}$

$2 =(1.011)^{t}$

Now let us solve for t using logarithm.

$ln(2) =ln (1.011)^{t}$

$ln(2) =t \cdot ln(1.011)$

$0.6931471805599453 =t \cdot 0.0109399400383344$

$t=\frac{0.6931471805599453}{0.0109399400383344}$

$t=63.3593217267282743049\approx 63.36$

Therefore, it will take 63.36 years the population to be double.

B) Now we will find the number of years it will take the population to be triple of its size.

$7.1 \times 3=7.1(1.011)^{t}$

$\frac{7.1 \times 3}{7.1} =(1.011)^{t}$

$3 =(1.011)^{t}$

Now let us solve for t using logarithm.

$ln(3) =ln (1.011)^{t}$

$ln(3) =t \cdot ln(1.011)$

$1.0986122886681097 =t \cdot 0.0109399400383344$

$t=\frac{1.0986122886681097}{0.0109399400383344}$

$t=100.4221490079915311298\approx 100.42$

Therefore, it will take 100.42 years the population to triple of its size.

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2 years ago

Joey and Armando live on the same st as the park. The park is 9/10 mile from Joey's home. Joey leaves home and walks to Armando'

AleksandrR [38]

Let's start by assuming Armando's house is between Joey's and the park.

Let $x$ be the distance Joey walked to Armando's house.

<span>The park is 9/10 mile from Joey's home. Joey leaves home and walks to Armando's home. Then Joey and Armando walk 3/5 mile to the park.

</span> $\dfrac{9}{10} = x + \dfrac{3}{5}$

$x = \dfrac{9}{10} - \dfrac{3}{5} = \dfrac{9}{10} -\dfrac{6}{10} = \dfrac{3}{10}$

That's probably the answer they're looking for. But what if the park is between Joey and Armando's houses or Joey is between the park and Armando? (The latter isn't really possible with the given distances.)

Let $a, b, c$ be the distances between three collinear points like we have here. Our equation is really a few equations in one, something like

$\pm a \pm b = \pm c$

Let's get rid of the plus/minuses. Squaring,

$a^2 + b^2\pm 2ab = c^2$

$\pm 2ab = c^2-a^2-b^2$

$4a^2b^2 = (c^2-a^2-b^2)^2$

For us, that's a quadratic equation for $c^2$

$4(9/10)^2(3/5)^2= (c^2-(9/10)^2 - (3/5)^2)^2$

I'll skip right to the solutions,

$c^2=\dfrac{9}{100} \textrm{ or } c^2=\dfrac{9}{4}$

$c=\dfrac{3}{10} \textrm{ or } c=\dfrac{3}{2}$

We could have gotten the 3/2 just by adding 9/10+3/5 but this was more fun.

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3 years ago