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aivan3 [116]
2 years ago
8

Awnser please now !!!!!!!!!!!!!!

Mathematics
2 answers:
polet [3.4K]2 years ago
5 0

Answer:

(2, 0)

Step-by-step explanation:

The x intercept is when y is equal to 0.

Therefore,

8x-1/3*0=16

8x=16

x=2

The answer is (2,0)

makkiz [27]2 years ago
4 0

Answer:

2,0

Step-by-step explanation: Sorry I don't have an step-by-step explanation .

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Find the volume of a pyramid with a square base , where the area of the base is 7.5 cm ^ 2 and the height of the pyramid is 8.6
svetlana [45]

Answer: 21.5cm³

Step-by-step explanation: This is your answer there is no way it is incorrect

4 0
2 years ago
I don't understand this and I can only miss 1 question on my quiz and already did​
Dahasolnce [82]

Answer:

23

Step-by-step explanation:

they add five

9+5=14

12+5=17

15+5=20

18+5=23

7 0
2 years ago
Read 2 more answers
The output is eight more than the input
d1i1m1o1n [39]

Answer:

x = input

y = output

y = 8 + x

7 0
3 years ago
Solve the following equation: |7 – (24 ÷ | 3-6 |)|<br>(A)15(B)-12(C)1(D)-1
Zepler [3.9K]
PEDMAS

The rule is Paranthesis First

|7-(24÷|3-6|)|

|7-(24÷|-3|)|

Now we have to get the -3 outside the 2 Lines but the lines are saying that if the number inside is negative when you get it outside it has to be positive.

|7-(24÷3)|-

|7-(8)|

|7-8|

|-1|

= 1

3 0
3 years ago
Read 2 more answers
Find the general solution of the following ODE: y' + 1/t y = 3 cos(2t), t &gt; 0.
Margarita [4]

Answer:

y = 3sin2t/2 - 3cos2t/4t + C/t

Step-by-step explanation:

The differential equation y' + 1/t y = 3 cos(2t) is a first order differential equation in the form y'+p(t)y = q(t) with integrating factor I = e^∫p(t)dt

Comparing the standard form with the given differential equation.

p(t) = 1/t and q(t) = 3cos(2t)

I = e^∫1/tdt

I = e^ln(t)

I = t

The general solution for first a first order DE is expressed as;

y×I = ∫q(t)Idt + C where I is the integrating factor and C is the constant of integration.

yt = ∫t(3cos2t)dt

yt = 3∫t(cos2t)dt ...... 1

Integrating ∫t(cos2t)dt using integration by part.

Let u = t, dv = cos2tdt

du/dt = 1; du = dt

v = ∫(cos2t)dt

v = sin2t/2

∫t(cos2t)dt = t(sin2t/2) + ∫(sin2t)/2dt

= tsin2t/2 - cos2t/4 ..... 2

Substituting equation 2 into 1

yt = 3(tsin2t/2 - cos2t/4) + C

Divide through by t

y = 3sin2t/2 - 3cos2t/4t + C/t

Hence the general solution to the ODE is y = 3sin2t/2 - 3cos2t/4t + C/t

3 0
3 years ago
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