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2 years ago

What is the part and whole of 40% of 20?

Mathematics

1 answer:

pashok25 [27]2 years ago

5 0

The answer is 8

I hope this helps!

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Write the quadratic equation whose roots are 3 and -5, and whose leading coefficient is 2.

Rainbow [258]

$\begin{cases} x=3\implies &x-3=0\\ x=-5\implies &x+5=0 \end{cases}\qquad \implies\qquad a(x-3)(x+5)=\stackrel{0}{y} \\\\\\ a(\stackrel{F~O~I~L}{x^2+2x-15})=y\implies \stackrel{\textit{leading coefficient of 2}}{2(x^2+2x-15)}=y\implies 2x^2+4x-30=y$

Check the picture below.

5 0

2 years ago

What is the length of the unknown leg of the right triangle?

Alik [6]

1. Use the Pythagorean Theorem since this is a right triangle:

a^2 + b^2 = c^2

3^2 + b^2 = 4^2

b^2 = 4^2 - 3^2

b = √ 7

b = 3

answer : 3

8 0

2 years ago

A line is defined by the equation 2 x + y = 4. Which shows the graph of this line?

Black_prince [1.1K]

Answer:

Step-by-step explanation:

divide by 4

$\frac{x}{2} +\frac{y}{4} =1\\x-intercept=2\\y-intercept=4\\draw~ a~ line~ through~ (2,0)~ and ~(0,4)$

it is the reqd. graph.

3 0

2 years ago

Read 2 more answers

What is the simplified form of (3.25x10^3)(7.8x10^6) written in scientific notation?

BabaBlast [244]

Answer: $2.535*10^{10}$

Step-by-step explanation:

1. By definition, scientific notation has the following form:

$a*10^{b}$

Where:

<em> a</em> is a number between 1 and 10 but does not include 10.

<em> b </em>an integer.

2. When you multiply 3.25 by 7.8 the result is 25.35 and when you multiply $10^{3}$ by $10^{6}$ , you obtain $10^{9}$ , but to write ree thsult in scientific notation, the decimal point must be after the first digit. Therefore, you must move the decimal point one place to the left. So, the exponent changes from 9 to 10:

$(3.25*10^{3})(7.8*10^{6})=2.535*10^{10}$

3 0

2 years ago

Factor completely, then place the answer in the proper location on the grid.

Tresset [83]

$\mathrm{Factor\:out\:common\:term\:}9 \ \textgreater \ 9\left(x^4-25y^8\right)$

$Factor \ \textgreater \ x^4-25y^8$

$\mathrm{Apply\:difference\:of\:squares\:rule:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)$
$x^2-5y^4=\left(x+\sqrt{5}y^2\right)\left(x-\sqrt{5}y^2\right)$

$x^2-5y^4 \ \textgreater \ \mathrm{Apply\:difference\:of\:squares\:rule:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)$
$x^2-5y^4=\left(x+\sqrt{5}y^2\right)\left(x-\sqrt{5}y^2\right)$

$\left(x^2+5y^4\right)\left(x+\sqrt{5}y^2\right)\left(x-\sqrt{5}y^2\right)$

$9\left(x^2+5y^4\right)\left(x+\sqrt{5}y^2\right)\left(x-\sqrt{5}y^2\right)$

Hope this helps!

8 0

2 years ago