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3 years ago

What’s da suace I need the awser

Mathematics

2 answers:

Aleksandr [31]3 years ago

8 0

Answer:

Step-by-step explanation:

-1, 3

LiRa [457]3 years ago

5 0

Answer:

y= -1x+ 3

i hope this helps

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64 POINTS

Mekhanik [1.2K]

L(t)=9-0.5t
t=time in hours,
L(t)=length of candle at time t

(A) substitute t=1,2,3,4,5,6 respectively to get
(1,8.5)
(2,8)
(3,7.5)
(4,7)
(5,6.5)
(6,6)

(B) it is a function because none of the x-values duplicate, (nor the y-values).

(C) If the rate changes to 0.45 "/hr, the relation is still linear.
All linear function with finite slope are functions.

7 0

3 years ago

Read 2 more answers

Is there an ordered pair solutions to -3x+3y=4 and -x+y=3 ????

elixir [45]

Y=x+3
-3x+3(x+3)=4
-3x+3x+9=4
9=4
Since this statement is false, there are no ordered pair solutions to this problem.

8 0

3 years ago

Please help me with this problem!

Oksanka [162]

(25-y) ^2-y^2=25
y=12
x=25-12
x=13

(13,12)

7 0

3 years ago

14, 8, 17, 21, x, 11, 3, 13; range: 20

MakcuM [25]

Answer:

x = 1

Step-by-step explanation:

When you subtract the smallest number from the biggest number, the answer is called the range. Here the largest number is 21. The smallest is 3. Then the range should be 18. But since it's 20. This mean that x is 1. Because 21 - 1 will equal 20, making 20 the range. So the correct answer is 1.

6 0

3 years ago

Evaluate the interval (Calculus 2)

Darya [45]

Answer:

$2 \tan (6x)+2 \sec (6x)+\text{C}$

Step-by-step explanation:

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{12}{1-\sin (6x)}\:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\$\displaystyle \int a\:\text{f}(x)\:\text{d}x=a \int \text{f}(x) \:\text{d}x$\end{minipage}}$

If the terms are multiplied by constants, take them outside the integral:

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)}\:\:\text{d}x$

Multiply by the conjugate of 1 - sin(6x) :

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)} \cdot \dfrac{1+\sin(6x)}{1+\sin(6x)}\:\:\text{d}x$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{1-\sin^2(6x)} \:\:\text{d}x$

$\textsf{Use the identity} \quad \sin^2 x+ \cos^2 x=1:$

$\implies \sin^2 (6x) + \cos^2 (6x)=1$

$\implies \cos^2 (6x)=1- \sin^2 (6x)$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

Expand:

$\implies 12\displaystyle \int \dfrac{1}{\cos^2(6x)}+\dfrac{\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

$\textsf{Use the identities }\:\: \sec \theta=\dfrac{1}{\cos \theta} \textsf{ and } \tan\theta=\dfrac{\sin \theta}{\cos \theta}:$

$\implies 12\displaystyle \int \sec^2(6x)+\dfrac{\tan(6x)}{\cos(6x)} \:\:\text{d}x$

$\implies 12\displaystyle \int \sec^2(6x)+\tan(6x)\sec(6x) \:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}$

$\boxed{\begin{minipage}{6 cm}\underline{Integrating $ \sec kx \tan kx$}\\\\$\displaystyle \int \sec kx \tan kx\:\text{d}x= \dfrac{1}{k}\sec kx\:\:(+\text{C})$\end{minipage}}$

$\implies 12 \left[\dfrac{1}{6} \tan (6x)+\dfrac{1}{6} \sec (6x) \right]+\text{C}$

Simplify:

$\implies \dfrac{12}{6} \tan (6x)+\dfrac{12}{6} \sec (6x)+\text{C}$

$\implies 2 \tan (6x)+2 \sec (6x)+\text{C}$

Learn more about indefinite integration here:

brainly.com/question/27805589

brainly.com/question/28155016

3 0

2 years ago