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Elis [28]
3 years ago
10

Please help due ASAP I really need help

Mathematics
2 answers:
elena55 [62]3 years ago
6 0
C because it is icoceles.
Oxana [17]3 years ago
4 0

Answer:

21.25

Step-by-step explanation:

The sum of all angles in a triangle equals 180°

Since it's an isosceles triangle, the base angles are congruent.

180 = 75 + (2x + 10) + (2x + 10)

180 = 95 + 4x

-95    -95

85 = 4x

x = 21.25

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Change 3/8 to a decimal fraction. 0.375 37.5 375 0.0375
meriva
3/8 as a decimal would be 0.375
3 0
3 years ago
Read 2 more answers
Which phrases can be used to represent the inequality 12-3x≥9? Check all that apply. (multi choice)
balandron [24]

Answer:

Which phrases can be used to represent the inequality 12-3x≥9? Check all that apply. (multi choice)

<u> A Twelve less than three times a number is greater than or equal to nine. </u>

B Twelve less than three times a number is at least nine.

C Twelve less three times a number is at least nine.

D Twelve minus the product of three and a number is no more than nine.

E Twelve less three times a number is at most nine.

F Twelve minus the product of three and a number is no smaller than nine.

G Twelve minus the product of three and a number is larger than nine.

Step-by-step explanation:

12 - 3x>=9

8 0
3 years ago
5.5-3b=2b-6.25 how do you slove
olga55 [171]
Add 6.25 from 5 then you get 11.25. then you add 3b to 2b then you get 5b. the. divide 5b to 11.25 to get the answer
3 0
4 years ago
Solve the inequality 4t^2 ≤ 9t-2 please show steps and interval notation. thank you!​
FromTheMoon [43]

Answer: t\in [\dfrac{1}{4},2]

Step-by-step explanation:

Given

Inequality is 4t^2\leq9t-2

Taking variables one side

\Rightarrow 4t^2-9t+2\leq0\\\Rightarrow 4t^2-8t-t+2\leq0\\\Rightarrow 4t(t-2)-1(t-2)\leq0\\\Rightarrow (4t-1)(t-2)\leq0

Using wavy curve method

t\in [\dfrac{1}{4},2]

3 0
3 years ago
(5) Find the Laplace transform of the following time functions: (a) f(t) = 20.5 + 10t + t 2 + δ(t), where δ(t) is the unit impul
Aloiza [94]

Answer

(a) F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}

(b) F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}

Step-by-step explanation:

(a) f(t) = 20.5 + 10t + t^2 + δ(t)

where δ(t) = unit impulse function

The Laplace transform of function f(t) is given as:

F(s) = \int\limits^a_0 f(s)e^{-st} \, dt

where a = ∞

=>  F(s) = \int\limits^a_0 {(20.5 + 10t + t^2 + d(t))e^{-st} \, dt

where d(t) = δ(t)

=> F(s) = \int\limits^a_0 {(20.5e^{-st} + 10te^{-st} + t^2e^{-st} + d(t)e^{-st}) \, dt

Integrating, we have:

=> F(s) = (20.5\frac{e^{-st}}{s} - 10\frac{(t + 1)e^{-st}}{s^2} - \frac{(st(st + 2) + 2)e^{-st}}{s^3}  )\left \{ {{a} \atop {0}} \right.

Inputting the boundary conditions t = a = ∞, t = 0:

F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}

(b) f(t) = e^{-t} + 4e^{-4t} + te^{-3t}

The Laplace transform of function f(t) is given as:

F(s) = \int\limits^a_0 (e^{-t} + 4e^{-4t} + te^{-3t} )e^{-st} \, dt

F(s) = \int\limits^a_0 (e^{-t}e^{-st} + 4e^{-4t}e^{-st} + te^{-3t}e^{-st} ) \, dt

F(s) = \int\limits^a_0 (e^{-t(1 + s)} + 4e^{-t(4 + s)} + te^{-t(3 + s)} ) \, dt

Integrating, we have:

F(s) = [\frac{-e^{-(s + 1)t}} {s + 1} - \frac{4e^{-(s + 4)}}{s + 4} - \frac{(3(s + 1)t + 1)e^{-3(s + 1)t})}{9(s + 1)^2}] \left \{ {{a} \atop {0}} \right.

Inputting the boundary condition, t = a = ∞, t = 0:

F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}

3 0
3 years ago
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