All categories

3 years ago

What is the value of-1/2(x-6) when x=2?

Mathematics

1 answer:

matrenka [14]3 years ago

7 0

The answer is 2.

-1/2(2-6)

-1/2(-4)=2

You might be interested in

17X+1 +20x-14 = 90 find x

professor190 [17]

Answer:

x = 2.8

Step-by-step explanation:

17x+1+20x-14 = 90

Combine like terms.

37x-13 = 90

Add 13 on both sides.

37x-13 = 90

+13 +13

37x = 103

Divide both sides by 37.

37x = 103

/37 /37

x = 2.8

(I rounded up, but if you need a more specific number, just search up 103 divided by 37.)

Hopefully this helps you!! Have an amazing day c:

7 0

3 years ago

What is .3125 expresses as a fraction???

d1i1m1o1n [39]

.3125 expressed as a fraction is 5/16

6 0

3 years ago

Walter pays $2.53 to ship a package

Ivan

Answer:

what is the rest of the equation?

Step-by-step explanation:

7 0

2 years ago

Given y = 3x + 1 State the quadrants in which this graph is in. (Use the numbers 1-4)

FrozenT [24]

I have no clue man or woman

3 0

3 years ago

On a cold day, hailstones fall with a velocity of 2i − 6k m s−1 . If a cyclist travels through the hail at 10i m s−1 , what is t

Paraphin [41]

Answer:

The velocity of the hail relative to the cyclist is $V_{hc}=-8\hat{i}-6\hat{k}$

The angle at which hailstones falling relative to the cyclist is $\theta = 36.86^\circ$

Step-by-step explanation:

Given : On a cold day, hailstones fall with a velocity of $2\hat{i}-6\hat{k}\text{ m/s}$ . If a cyclist travels through the hail at $10\hat{i} \text{ m/s}$ .

To find : What is the velocity of the hail relative to the cyclist and At what angle are the hailstones falling relative to the cyclist?

Solution :

The velocity of the hailstone falls is $V_h=2\hat{i}-6\hat{k}\text{ m/s}$

The velocity of the cyclist travels through the hail is $V_c=10\hat{i} \text{ m/s}$

The velocity of the hail relative to the cyclist is given by,

$V_{hc}=V_h-V_c$

Substitute the value in the formula,

$V_{hc}=2\hat{i}-6\hat{k}-10\hat{i}$

$V_{hc}=-8\hat{i}-6\hat{k}$

So, The velocity of the hail relative to the cyclist is $V_{hc}=-8\hat{i}-6\hat{k}$

Now, The angle of hails falling relative to the cyclist is given by

$\theta = \tan^{-1}(\frac{-6}{-8})$

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

$\theta = 36.86^\circ$

So, The angle at which hailstones falling relative to the cyclist is $\theta = 36.86^\circ$

4 0

3 years ago