Ask question

Ask question

All categories

3 years ago

15

4 x ½ what is the answer

1 answer:

Strike441 [17]3 years ago

7 0

Answer:

2

Step-by-step explanation:

4/1×1/2 =4×1

1 x2 =4/2 =2

<h2><u><em>Hope it helps</em></u></h2>

You might be interested in

PLEASE SOLVE AND CHECK. SHOW COMPLETE SOLUTION

Alex17521 [72]

<u>Solution</u><u>:</u>

$\sqrt{4x + 13} = x + 2$

First square both sides.

$= > ( \sqrt{4x + 13} ) ^{2} = (x + 2) ^{2}$

Now, square root and square gets cancel out in the LHS. And in the RHS, apply the identity: (a + b)² = a² + 2ab + b².

$= > 4x + 13 = {(x)}^{2} + 2 \times x \times 2 + (2) ^{2} \\ = > 4x + 13 = {x}^{2} + 4x + 4$

Now, transpose 4x and 4 to LHS.

$= > 4x - 4x + 13 - 4 = {x}^{2} \\$

Now, do the addition and subtraction.

$= > {x}^{2} = 9 \\ = > x = \sqrt{9} \\ = > x = ±3$

<u>Answer</u><u>:</u>

<u>x </u><u>=</u><u> </u><u>±</u><u> </u><u>3</u>

Hope you could understand.

If you have any query, feel free to ask.

3 0

2 years ago

At the Arctic weather station, a warning light turns on if the outside temperature is below -25 degrees Fahrenheit. Which inequa

Leya [2.2K]

Answer:

t≥-25

Step-by-step explanation:

this is becuaset ≥ -25 shows that it can not fall under -25, but can be equal to -25.

8 0

2 years ago

Oml<br> 1/6cm<br> 4cm<br> What the surface area of this shape

Kitty [74]

1/6 x 4 is .666666...

6 0

3 years ago

A cyclist travels at distance of 400 meters in 120 seconds towards school, calculate his speed. (Show your work)

DochEvi [55]

we know that

The scalar magnitude of the velocity vector is the speed. The speed is equal to

$Speed=\frac{distance}{time}$

in this problem we have

$distance=400\ m \\time=120\ sec$

substitute in the formula

$Speed=\frac{400}{120}$

$Speed=3.5\frac{m}{sec}$

therefore

<u>the answer Part a) is</u>

the speed is equal to $3.5\frac{m}{sec}$

<u>Part b) </u>Find the velocity

we know that

<u>Velocity </u>is a vector quantity; both magnitude and direction are needed to define it

in this problem we have

the magnitude is equal to the speed

$magnitude=3.5\frac{m}{sec}$

$direction=North\ East\ (NE)$

therefore

<u>the answer Part b) is</u>

the velocity is $3.5\frac{m}{sec}\ North\ East\ (NE)$

Part c)

we know that

the acceleration is equal to the formula

$a=\frac{V2-V1}{t2-t1}$

in this problem we have

$V2=0 \\V1=3.5\frac{m}{sec}$

$t2=15\ sec\\t1=0$

substitute in the formula

$a=\frac{0-3.5}{15-0}$

$a=-\frac{3.5}{15}\frac{m}{sec^{2}}$

$a=-\frac{7}{30}\frac{m}{sec^{2}}$

therefore

<u>the answer Part c) is</u>

the acceleration is $-\frac{7}{30}\frac{m}{sec^{2}}$

This is an example of negative acceleration

7 0

3 years ago

Read 2 more answers

If W(-10, 4), X(-3, -1), and Y(-5, 11) classify ΔWXY by its sides. Show all work to justify your answer.

solniwko [45]

Given:

The vertices of ΔWXY are W(-10, 4), X(-3, -1), and Y(-5, 11).

To find:

Which type of triangle is ΔWXY by its sides.

Solution:

Distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula, we get

$WX=\sqrt{(-3-(-10))^2+(-1-4)^2}$

$WX=\sqrt{(-3+10)^2+(-5)^2}$

$WX=\sqrt{(7)^2+(-5)^2}$

$WX=\sqrt49+25}$

$WX=\sqrt{74}$

Similarly,

$XY=\sqrt{\left(-5-\left(-3\right)\right)^2+\left(11-\left(-1\right)\right)^2}=2\sqrt{37}$

$WY=\sqrt{\left(-5-\left(-10\right)\right)^2+\left(11-4\right)^2}=\sqrt{74}$

Now,

$WX=WY$

So, triangle is an isosceles triangles.

and,

$WX^2+WY^2=(\sqrt{74})^2+(\sqrt{74})^2$

$WX^2+WY^2=74+74$

$WX^2+WY^2=148$

$WX^2+WY^2=(2\sqrt{37})^2$

$WX^2+WY^2=WY^2$

So, triangle is right angled triangle.

Therefore, the ΔWXY is an isosceles right angle triangle.

3 0

2 years ago