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3 years ago

Pls help with this all questions

Mathematics

1 answer:

Orlov [11]3 years ago

7 0

Look you math book up on sladder and it'll have all the answers for free!!!!!!!!!!!!!!!

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Pls help ASAP need in 10 mins

Rina8888 [55]

Answer:

Step-by-step explanation:

i think its C sorry if its wrong

8 0

3 years ago

Solve the following quadratic equation by completing the square: x² + 12x = 15

Lerok [7]

Answer:

x = -1.14 or -13.14

Explanations:

The given equation is:

$x^2+12x\text{ = 15}$

Find the half of 12, and add the square to both sides of the equation.

That is add 6² to both sides

$x^2+12x\text{ + 6}^{2}\text{ = 15 + 6}^{2}$ $\begin{gathered} (x+6)^2\text{ = 15 + 3}6 \\ (x+6)^2\text{ = 51} \\ \end{gathered}$

Find the square root of both sides:

$\begin{gathered} \sqrt[]{(x+6)^2_{}}=\pm\sqrt[]{51} \\ \text{x + 6 = }\pm\sqrt[]{51} \\ x\text{ = -6 }\pm\sqrt[]{51} \\ \text{x = }-6\pm7.14 \\ x\text{ = -6+7.14 = }1.14 \\ x\text{ = -6 - 7.14} \\ \text{x = -13.14} \end{gathered}$

x = -1.14 or -13.14

3 0

11 months ago

A cereal manufacturer decides to offer a new family-sized box based on the regular-sized box. They want the volume of the family

Pani-rosa [81]

The answer would be an increase by a factor of 3/2

4 0

2 years ago

Read 2 more answers

Could you help me pleas

tia_tia [17]

21 for 3 and 10 for 70

8 0

2 years ago

Read 2 more answers

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.

3 0

2 years ago