X - 4y - 1=0 x + 5y - 4= 0

Mathematics

1 answer:

Alex17521 [72]3 years ago

3 0

Would you rather elimination or substitution

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What does this problem equal?show your work.<br><br>-12(k+4)=60

HACTEHA [7]

Hey there!

First, divide both sides by 12
-12(k+4)/-12= 60/-12 and you would getk+4=5
Now subtract 4 from both sides

k+4-4= 5-4

k=-9

3 0

3 years ago

Read 2 more answers

HELP ME PLEASE I DONT KNOW HOW TO DO THIS

Diano4ka-milaya [45]

Answer:

x = - 6

p = 9

k = 1

Step-by-step explanation:

See attached worksheet.

4 0

2 years ago

Find the derivative.

Aleksandr [31]

Answer:

Using either method, we obtain: $t^\frac{3}{8}$

Step-by-step explanation:

a) By evaluating the integral:

$\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du$

The integral itself can be evaluated by writing the root and exponent of the variable u as: $\sqrt[8]{u^3} =u^{\frac{3}{8}$

Then, an antiderivative of this is: $\frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}$

which evaluated between the limits of integration gives:

$\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}$

and now the derivative of this expression with respect to "t" is:

$\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}$

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

$g(x)=\int\limits^x_a {f(t)} \, dt$

is continuous on [a,b], differentiable on (a,b) and $g'(x)=f(x)$

Since this this function $u^{\frac{3}{8}$ is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means: