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3 years ago

Factor the expression 5x^2 -22x -15

Mathematics

1 answer:

Murljashka [212]3 years ago

3 0

$5x^2-22x-15=5x^2\underbrace{-25x+3x}_{-22x}-15\\\\=(5x^2-25x)+(3x-15)=5x(x-5)+3(x-5)\\\\=\boxed{(x-5)(5x+3)}$

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Answer:

A. 5.0 * 10⁻⁹ J

Step-by-step explanation:

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2 years ago

Find the area of a rhombus having perimeter 80 em and one diagonal 24 cm.<br>

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2 years ago

What are the domain and range of the function f(x)=3/4x+5?

dybincka [34]

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2 years ago

Robin has a certain number of dimes in his pocket. If the dimes were pennies, he would have $1.08 less than he has now. How many

laiz [17]

Answer:

Robin will have 12 dimes in his pocket.

Step-by-step explanation:

Since;

D = One dime = $0.1

P = One penny = $0.01

Amount to have less = $1.08

From the above, when a dime is changed to a penny, the amount held by Robin will decrease by $0.09. That is,

Decease in holding = D - P = $0.1 - $0.01 = $0.09

Therefore, we have:

Number of dimes = Amount to have less / Decease in holding = $1.08 / $0.09 = 12 dimes

Therefore, Robin will have 12 dimes in his pocket.

4 0

2 years ago

Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.

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2 years ago