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2 years ago

How to solve this question

Mathematics

1 answer:

vichka [17]2 years ago

3 0

#4 has the same y intercept
If you put the equations in y = mx + b you will see that it has a y intercept of -3

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When this 3-digit number is rounded to the nearest hundred, it rounds to 200. Rounded to the nearest ten, this number rounds to

alisha [4.7K]

199, it rounds to 200 because it's closer to 200 than 100. It is also closer to 200 than to 190. And finally, 1 + 9 + 9 = 19

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2 years ago

45 + 57 = x + y<br> x= 37<br> y =?

Mekhanik [1.2K]

Answer:

y = 65

Step-by-step explanation:

45 + 57 = x + y

x = 37

45 + 57 = 37 + y

45 + 57 = 102

102 = 37 + y

-37 -37

102 - 37 = 65

37 + y - 37 = y

65 = y

y = 65

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3 years ago

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The function f(x) = 2x2 + 3x + 5, when evaluated, gives a value of 19. What is the function’s input value?

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The function's input value is 2.

f(x) = 2x² + 3x + 5
f(x) = 2(2)² + 3(2) + 5
f(x) = 2(4) + 6 + 5
f(x) = 8 + 11
f(x) = 19

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3 years ago

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If b=3 what does -3b equal

Aleks04 [339]

-3b=-9. Plug in 3. -3(3)=-9.

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2 years ago

A car is moving with a velocity (3.0 m/s) x̂ + (1.0 m/s) ŷ and 3.0 seconds later its velocity is (6.0 m/s) x̂ - (3.0 m/s) ŷ. Wha

Mars2501 [29]

Here we know that the initial velocity of the car is given by:

$V_i=(3.0\hat x+1.0\hat y)m/s$

And the final velocity of the car is given by:

$V_f=(6.0\hat x-3.0\hat y)m/s$

It took 3 seconds to attain the final velocity, so we have $t=3 s$

Therefore, the acceleration can be obtained by:

$V_f-V_i=a\times t$

$a=\frac{V_f-V_i}{t}$

Plugging the values of the initial, final velocity and the time, we get:

$a= \frac{(6\hat x-3\hat y)-(3 \hat x+1 \hat y)}{3} =\frac{6\hat x-3 \hat y -3 \hat x-1 \hat y}{3} =\frac{3 \hat x-4 \hat y}{3}$

So the acceleration of the car is given by:

$a=\frac{3 \hat x}{3} -\frac{4 \hat y}{3} =1 \hat x- \frac{4}{3} \hat y$

Now we need to find the direction of the average acceleration of the car:

$\theta=tan^{-1} \frac{y}{x}$

Here, x and y are the coefficients of the 'x' and 'y' components of the vector:

$\theta=tan^{-1}(\frac{-\frac{4}{3} }{1} )=tan^{-1}(\frac{-4}{3}) =-53.13^\circ$

Therefore, the direction of the average acceleration of the car is $-53.13^\circ$ .

8 0

3 years ago