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3 years ago

220/(divide) by 31.25 step by step or write in paper and send

Mathematics

1 answer:

diamong [38]3 years ago

4 0

Answer:

7.04

Step-by-step explanation:

The computation is shown below:

Given that

220 would be divided by 31.25

i.e.

= 220 ÷ 31.25

= 7.04

After dividing the 220 from 31.25 it provides an answer i.e. 7.04

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Point P was rotated about the Irgun (0,0) by -15

mart [117]

Answer:

Step-by-step explanation:

The point you're looking for is the same distance from the origin, but 15° clockwise around a circle centered at the origin. Points A and D are the only ones with the same radius. Point D is the only one that is 15° clockwise from P.

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The approximate coordinates of the rotated point are (-4.3120, 3.2259).

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3 years ago

A line goes through the point (6,-2) and has a slope of -3. What is the value of a if the point (a,7) lies on the line?

Vikki [24]

Answer:

Step-by-step explanation:

The slope of a line given two points can be found by using the formula

$m = \frac{ y_2 - y _ 1}{x_ 2 - x_ 1} \\$

From the question the slope is - 3

To find the value of a we have

$- 3 = \frac{7 - - 2}{a - 6} \\ - 3 = \frac{7 + 2}{a - 6} \\ - 3(a - 6) = 9 \\ - 3a + 18 = 9 \\ - 3a = 9 - 18 \\ - 3a = - 9$

Divide both sides by - 3

$\frac{ - 3a}{ - 3} = \frac{ - 9}{ - 3} \\$

We have the final answer as

Hope this helps you

3 0

3 years ago

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How would I solve 16x^4-41x^2+25=0 ???

sveticcg [70]

$16{ x }^{ 4 }-41{ x }^{ 2 }+25=0$

${ x }^{ 4 }={ ({ x }^{ 2 }) }^{ 2 }\\ \\ 16{ ({ x }^{ 2 }) }^{ 2 }-41{ x }^{ 2 }+25=0$

First of all to make our equation simpler, we'll equal $x^{2}$ to a variable like 'a'.

So,

${ x }^{ 2 }=a$

Now let's plug $x^{2}$ 's value (a) into the equation.

$16{ ({ x }^{ 2 }) }^{ 2 }-41{ x }^{ 2 }+25=0\\ \\ { x }^{ 2 }=a\\ \\ 16{ (a) }^{ 2 }-41{ a }+25=0$

Now we turned our equation into a quadratic equation.

(The variable 'a' will have a solution set of two solutions, but 'x' , which is the variable of our first equation will have a solution set of four solutions since it is a quartic equation (<span>fourth-degree <span>equation) )

Let's solve for a.

The formula used to solve quadratic equations ;

$\frac { -b\pm \sqrt { { b }^{ 2 }-4\cdot t\cdot c } }{ 2\cdot t }$

The formula is used in an equation formed like this :
</span></span>
$t{ x }^{ 2 }+bx+c=0$

In our equation,

t=16 , b=-41 and c=25

Let's plug the values in the formula to solve.

$t=16\quad b=-41\quad c=25\\ \\ \frac { -(-41)\pm \sqrt { -(41)^{ 2 }-4\cdot 16\cdot 25 } }{ 2\cdot 16 } \\ \\ \frac { 41\pm \sqrt { 1681-1600 } }{ 32 } \\ \\ \frac { 41\pm \sqrt { 81 } }{ 32 } \\ \\ \frac { 41\pm 9 }{ 32 }$

So the solution set :

$\frac { 41+9 }{ 32 } =\frac { 50 }{ 32 } \\ \\ \frac { 41-9 }{ 32 } =\frac { 32 }{ 32 } =1\\ \\ a\quad =\quad \left\{ \frac { 50 }{ 32 } ,\quad 1 \right\}$

We found a's value.

Remember,

${ x }^{ 2 }=a$

So after we found a's solution set, that means.

${ x }^{ 2 }=\frac { 50 }{ 32 }$

and

${ x }^{ 2 }=1$

We'll also solve this equations to find x's solution set :)

${ x }^{ 2 }=\frac { 50 }{ 32 } \\ \\ \frac { 50 }{ 32 } =\frac { 25 }{ 16 } \\ \\ { x }^{ 2 }=\frac { 25 }{ 16 } \\ \\ \sqrt { { x }^{ 2 } } =\sqrt { \frac { 25 }{ 16 } } \\ \\ x=\quad \pm \frac { 5 }{ 4 }$

${ x }^{ 2 }=1\\ \\ \sqrt { { x }^{ 2 } } =\sqrt { 1 } \\ \\ x=\quad \pm 1$

So the values x has are :

$\frac { 5 }{ 4 }$ , $-\frac { 5 }{ 4 }$ , $1$ and $-1$

Solution set :

$x=\quad \left\{ \frac { 5 }{ 4 } \quad ,\quad -\frac { 5 }{ 4 } \quad ,\quad 1\quad ,\quad -1 \right\}$

I hope this was clear enough. If not please ask :)

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3 years ago

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(f +g)x= g(x)+ f (x)<br><br> Is the this true or false?

Bess [88]

Answer:

True

Step-by-step explanation:

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