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julia-pushkina [17]

3 years ago

5

Two cars start from the same point on flat ground. The first car drives off directly North 25 ms^-1. Two seconds later the secon

d car drives off directly East at 20 ms.^-1 After how many seconds are they 1 km apart?

1 answer:

ale4655 [162]3 years ago

8 0

Answer:

1)jungkook

2)jimin

3)jin

4)RM

5)Jhope

6)suga

7) V

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Which is larger 3kL or 30000L

dolphi86 [110]

30000L is larger than 3kL

7 0

2 years ago

Consider the following pair of equations:

Zinaida [17]

Answer:

The lines representing these equations intercept at the point (-4,2) on the plane.

Step-by-step explanation:

When we want to find were both lines intercept, we are trying to find a pair of values (x,y) that belongs to both equations, which means that it satisfies both equations at the same time.

Therefore, we can use the second equation that gives us the value of y in terms of x, to substitute for y in the first equation. Then we end up with an equation with a unique unknown, for which we can solve:

$x+y=-2\\x+(2x+10)=-2\\x+2x+10=-2\\3x+10=-2\\3x=-2-10\\3x=-12\\x=\frac{-12}{3} \\x=-4$

Next we use this value we obtained for x (-4) in the same equation we use for substitution in order to find which y value corresponds to this:

$y=2x+10\\y=2(-4)+10\\y=-8+10\\y=2$

Then we have the pair (x,y) that satisfies both equations (-4,2), which is therefore the point on the plane where both lines intercept.

5 0

3 years ago

Use a calculator to find the approximate value of sin-1( 0.635).

Ne4ueva [31]

Using a calculator, I arrived to answer C because if you enter it in right you arrive at 39.419, rounding it to 39.42

8 0

3 years ago

Read 2 more answers

The sum of squares of two consecutive positive even integers is 52 more than their product. Find these numbers.

natulia [17]

Answer:

The first set of consecutive even integers equals (8 , 6)

The second set is ( - 8 and - 6) which also works.

Step-by-step explanation:

Equation

(x)^2 + (x + 2)^2 = (x)(x + 2) + 52 Remove the brackets on both sides

Solution

x^2 + x^2 + 4x + 4 = x^2 + 2x + 52 Collect the like terms on the left

2x^2+ 4x+ 4 = x^2 + 2x + 52 Subtract right side from left

2x^2 - x^2 + 4x - 2x + 4 - 52 = 0 Collect the like terms

x^2 + 2x - 48 = 0 Factor

(x + 8)(x - 6) = 0

Answer

Try the one you know works.

x - 6 = 0

x = 6

Therefore the two integers are 6 and 8

6^2 + 8^2 = 100

6*8 + 52 = 100

So 6 and 8 is one set of consecutive even numbers that works.

========================

What about the other set.

x + 8 = 0

x = - 8

x and x + 2

- 8 and -8 + 2 = - 8, - 6

(- 8 )^2 + (- 6)^2 = 100

(-8)(-6) + 52 = 100

Both sets of consecutive numbers work.

5 0

3 years ago

Read 2 more answers

Please someone help me to prove this..

Pachacha [2.7K]

Answer: see proof below

<u>Step-by-step explanation:</u>

Use the following Sum to Product Identities:

$\sin x+\sin y=2\sin \bigg(\dfrac{x+y}{2}\bigg)\cos \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\sin x-\sin y=2\cos \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x+\cos y=2\cos \bigg(\dfrac{x+y}{2}\bigg)\cos \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x+\cos y=-2\sin \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)$

<u>Proof LHS → RHS</u>

$\text{LHS:}\qquad \qquad \qquad \dfrac{\sin 5-\sin 15+\sin 25 - \sin 35}{\cos 5-\cos 15- \cos 25 + \cos 35}$

$\text{Reqroup:}\qquad \qquad \qquad \dfrac{(\sin 25+\sin 5)-(\sin 35 + \sin 15)}{(\cos 35+\cos 5)-(\cos 25 + \cos 15)}$

$\text{Sum to Product:}\quad \dfrac{2\sin \bigg(\dfrac{25+5}{2}\bigg)\cos \bigg(\dfrac{25-5}{2}\bigg)-2\sin \bigg(\dfrac{35+15}{2}\bigg)\cos \bigg(\dfrac{35-15}{2}\bigg)}{2\cos \bigg(\dfrac{25+15}{2}\bigg)\cos \bigg(\dfrac{25-15}{2}\bigg)-2\cos \bigg(\dfrac{35+5}{2}\bigg)\cos \bigg(\dfrac{35-5}{2}\bigg)}$ $\text{Simplify:}\qquad \qquad \dfrac{2\sin 15\cos 10-2\sin 25\cos 10}{2\cos 20\cos 15-2\cos 20\cos 5}$

$\text{Factor:}\qquad \qquad \dfrac{2\cos 10(\sin 15-\sin 25)}{2\cos 20(\cos 15-\cos 5)}$

$\text{Sum to Product:}\qquad \dfrac{\cos 10\bigg[2\cos \bigg(\dfrac{15+25}{2}\bigg)\sin \bigg(\dfrac{15-25}{2}\bigg)\bigg]}{\cos 20\bigg[-2\sin \bigg(\dfrac{15+5}{2}\bigg)\sin \bigg(\dfrac{15-5}{2}\bigg)\bigg]}$

$\text{Simplify:}\qquad \qquad \dfrac{\cos 10[2\cos 20\sin (-5)]}{\cos 20[-2\sin 10\sin 5]}\\\\\\.\qquad \qquad \qquad =\dfrac{-2\cos10 \cos 20 \sin 5}{-2\sin 10 \cos 20 \sin 5}\\\\\\.\qquad \qquad \qquad =\dfrac{\cos 10}{\sin 10}\\\\\\.\qquad \qquad \qquad =\cot 10$

LHS = RHS: cot 10 = cot 10 $\checkmark$

8 0

2 years ago