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4 years ago

X:2 (105-275:11):4=28

Mathematics

1 answer:

balandron [24]4 years ago

4 0

X:2+(105-275:11):4=28
X:2+(105-25):4=28
X:2+80:4=28
X:2+20=28
X:2=28-20
X:2=8
X=8×2
X=16

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What number is greater 5 or 5.01

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4 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 $\theta = 36.86^\circ$

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4 years ago

Multiple 3/4 times 16/9

nataly862011 [7]

3/4 times 16/9 is equal to 1 1/3.

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3 years ago

Read 2 more answers

1. Write the equation for a circle with center (1/2,3/7) and radius √13.

zhuklara [117]

Answer:

(x -1/2)^2 + (y -3/7)^2 = 13

Step-by-step explanation:

The equation for a circle with center (h, k) and radius r is ...

(x -h)^2 +(y -k)^2 = r^2

Filling in the given values results in the equation shown in the Answer space above.

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4 years ago