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3 years ago

The sum of 5 fifteens and 3 fifteens

Mathematics

2 answers:

Flauer [41]3 years ago

4 0

Answer:

(5 x 15) + (3 x 15) = 120

Step-by-step explanation:

5 x 15 = 75

3 x 15 = 45

75 + 45 = 120

Hope this helps :)

DedPeter [7]3 years ago

3 0

Answer:

5 * 15 = 75

3 * 15 = 45

45 + 75 = 120

Step-by-step explanation:

When you have a equation like that, you need to multiply. It is a faster way to solve the problem.

You had 15 five times.

15+15+15+15+15= 75 It equals the same thing because it is the same as 5 * 15

Same for the three fifteens.

15+15+15= 45.

The sum of the whole equation is 120 because you add 75 and 45.

Hope this helps, have a good day/night. Stay safe and healthy!

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A family tree business removes overgrown trees for 3 months in the summer. In June, they removed 1/3 the number of trees as they

devlian [24]

Answer: 299

Step-by-step explanation:

Given

Family removes one-third of tress in June as much as they dd in July

They remove twice the number of July in august

Suppose they removed x trees in July

for June it is, $\frac{x}{3}$

for August, it is $2x$

If they removed less than 1000 tress in 3 months

$\Rightarrow \dfrac{x}{3}+x+2x$

So, maximum number of tress removed in July is 299

7 0

3 years ago

Xavier is a salesperson who is paid a fixed amount of $455 per week. He also earns a commission of 3% on the sales he makes. If

melomori [17]

Xavier wants to earn more then 575 dollars a week, he will have to sale x<3,600 a week . So the answer is C

5 0

3 years ago

Plz help i dont know this i will mark brainliest if correct answer only

stiv31 [10]

Answer:

just double the number each time.

Step-by-step explanation:

so it would be 1, 2, 4, 8, 16, 32, 64, etc.

8 0

3 years ago

Read 2 more answers

Law of sines: StartFraction sine (uppercase A) Over a EndFraction = StartFraction sine (uppercase B) Over b EndFraction = StartF

Vsevolod [243]

First of all, this problem is properly done with the Law of Cosines, which tells us

$a^2 = b^2 + c^2 - 2 b c \cos A$

giving us a quadratic equation for b we can solve. But let's do it with the Law of Sines as asked.

$\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}$

We have c,a,A so the Law of Sines gives us sin C

$\sin C = \dfrac{c \sin A}{a} = \dfrac{5.4 \sin 20^\circ}{3.3} = 0.5597$

There are two possible triangle angles with this sine, supplementary angles, one acute, one obtuse:

$C_a = \arcsin(.5597) = 34.033^\circ$

$C_o = 180^\circ - C_a = 145.967^\circ$

Both of these make a valid triangle with A=20°. They give respective B's:

$B_a = 180^\circ - A - C_a = 125.967^\circ$

$B_o = 180^\circ - A - C_o = 14.033^\circ$

So we get two possibilities for b:

$b = \dfrac{a \sin B}{\sin A}$

$b_a = \dfrac{3.3 \sin 125.967^\circ}{\sin 20^\circ} = 7.8$

$b_o = \dfrac{3.3 \sin 14.033^\circ}{\sin 20^\circ} = 2.3$

Answer: 2.3 units and 7.8 units

Let's check it with the Law of Cosines:

$a^2 = b^2 + c^2 - 2 b c \cos A$

$0 = b^2 - (2 c \cos A)b + (c^2-a^2)$

There's a shortcut for the quadratic formula when the middle term is 'even.'

$b = c \cos A \pm \sqrt{c^2 \cos^2 A - (c^2-a^2)}$

$b = c \cos A \pm \sqrt{c^2( \cos^2 A - 1) + a^2}$

$b = 5.4 \cos 20 \pm \sqrt{5.4^2(\cos^2 20 -1) + 3.3^2}$

$b = 2.33958 \textrm{ or } 7.80910 \quad\checkmark$

Looks good.

6 0

3 years ago

Read 2 more answers

Reviewing Imaginary Numbers. Simplify I^12

melamori03 [73]

Answer:

Step-by-step explanation:

$i^{12}= i^{2} *i^{2} *i^{2} *i^{2} *i^{2} *i^{2}$

Remember that $i^{2}=-1$

Plug it in:

$i^{12}= (-1)^{2} *(-1)^{2} *(-1)^{2} *(-1)^{2} *(-1)^{2} *(-1)^{2}$

$=1*1*1*1*1*1$

$=1$

6 0

3 years ago