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3 years ago

PLEASE HELP!!! 100 POINTS !!!!!

Mathematics

1 answer:

bearhunter [10]3 years ago

4 0

Answer: if a triangle has congruent sides that means that the angles are also congruent.

Step-by-step explanation: It is not possible to change the angles of a triangle with congruent sides by rearranging them. All angles would remain the same degree.

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ITS URGENT (5−x)(2x+7)(x−3)≤0

tresset_1 [31]

Answer:

− 7 2 ≤ x ≤ 3 or x ≥ 5

Step-by-step explanation:

Basically isolate X,

(Sorry, I don't have time to explain deeply)

Hope I helped!! :)

7 0

3 years ago

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Solve C = 2πr (solve for "r") C = 15 How dodo solve it

Y_Kistochka [10]

Substitute C for 15 then divide 2(pi). (pi)= 3.14
15 / 2(3.14)
15 / 6.28
r = 2.39
I hope this helped love!!

4 0

4 years ago

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4x^2 y+8xy'+y=x, y(1)= 9, y'(1)=25

jarptica [38.1K]

Answer with explanation:

$\rightarrow 4x^2y+8x y'+y=x\\\\\rightarrow 8xy'+y(1+4x^2)=x\\\\\rightarrow y'+y\times\frac{1+4x^2}{8x}=\frac{1}{8}$

--------------------------------------------------------Dividing both sides by 8 x

This Integration is of the form ⇒y'+p y=q,which is Linear differential equation.

Integrating Factor

$=e^{\int \frac{1+4x^2}{8x} dx}\\\\e^{\log x^{\frac{1}{8}+\frac{x^2}{2}}\\\\=x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}$

Multiplying both sides by Integrating Factor

$x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\times [y'+y\times\frac{1+4x^2}{8x}]=\frac{1}{8}\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\\\\ \text{Integrating both sides}\\\\y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\frac{1}{8}\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx \\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=-[x^{\frac{9}{8}}]\times\frac{ \Gamma(0.5625, -x^2)}{(-x^2)^{\frac{9}{16}}}\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=(-1)^{\frac{-1}{8}}[ \Gamma(0.5625, -x^2)]+C-----(1)$

When , x=1, gives , y=9.

Evaluate the value of C and substitute in the equation 1.

6 0

3 years ago

It took 30 1-foot square tiles to cover the floor of the Henderson's entry. The length and width of their family room is three t

mixer [17]

Answer:

27 boxes

Step-by-step explanation:

Number of tiles required to cover the entry floor = 30

Since the size of each tile is 1-foot square, then;

area of Henderson's entry = area of each tile x 30

= 1 x 1 x 30

= 30 $ft^{2}$

Let the length of the entry be represented by x, and its width by y.

So that;

length of the family room = 3x

width of the family room = 3y

Area of Henderson's family room = length x width

= 3x x 3y

= 9xy

But,

xy = Area of Henderson's entry = 30 $ft^{2}$

So that;

Area of Henderson's family room = 9 x 30

= 270 $ft^{2}$

Therefore, 270 numbers of 1-foot square tiles would be required to tile the Henderson's family room.

Given that: 10 tiles come in a box number of box to tile the family room can be determined by;

number of box = $\frac{270}{10}$

= 27

The number of boxes required is 27.

7 0

3 years ago

List all the factors of 63

labwork [276]

The factors of 63 is 1,3,7,9,21,63

8 0

4 years ago

Read 2 more answers