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3 years ago

K=d d=5 find k Determine the value of the indicated variable

Mathematics

2 answers:

Igoryamba3 years ago

5 0

K=d d=5 find k well if d is 5 and k and d are equal then k=5

svet-max [94.6K]3 years ago

4 0

K=d
d=5

Since K is equal to d, and d is 5, the value of K is 5.

K=5
5(K)=5(d)

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Select the correct answer.

densk [106]

Answer:

Should be A

Step-by-step explanation:

6 0

3 years ago

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Round off 9.999 to the nearest 10

dlinn [17]

Answer:

the answer is 10 you round up a nine it keeps going to the end

6 0

3 years ago

In Exercises 45–48, let f(x) = (x - 2)2 + 1. Match the function with its graph

MA_775_DIABLO [31]

Answer:

45) The function corresponds to graph A

46) The function corresponds to graph C

47) The function corresponds to graph B

48) The function corresponds to graph D

Step-by-step explanation:

We know that the function f(x) is:

$f(x)=(x-2)^{2}+1$

45)

The function g(x) is given by:

$g(x)=f(x-1)$

using f(x) we can find f(x-1)

$g(x)=((x-1)-2)^{2}+1=(x-3)^{2}+1$

If we take the derivative and equal to zero we will find the minimum value of the parabolla (x,y) and then find the correct graph.

$g(x)'=2(x-3)$

$2(x-3)=0$

$x=3$

Puting it on g(x) we will get y value.

$y=g(3)=(3-3)^{2}+1$

$y=g(3)=1$

Then, the minimum point of this function is (3,1) and it corresponds to (A)

46)

Let's use the same method here.

$g(x)=f(x+2)$

$g(x)=((x+2)-2)^{2}+1$

$g(x)=(x)^{2}+1$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2x$

$0=2x$

$x=0$

Evaluatinf g(x) at this value of x we have:

$g(0)=(x)^{2}+1$

$g(0)=1$

Then, the minimum point of this function is (0,1) and it corresponds to (C)

47)

Let's use the same method here.

$g(x)=f(x)+2$

$g(x)=(x-2)^{2}+1+2$

$g(x)=(x-2)^{2}+3$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2(x-2)$

$0=2(x-2)$

$x=2$

Evaluatinf g(x) at this value of x we have:

$g(2)=(2-2)^{2}+3$

$g(2)=3$

Then, the minimum point of this function is (2,3) and it corresponds to (B)

48)

Let's use the same method here.

$g(x)=f(x)-3$

$g(x)=(x-2)^{2}+1-3$

$g(x)=(x-2)^{2}-2$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2(x-2)$

$0=2(x-2)$

$x=2$

Evaluatinf g(x) at this value of x we have:

$g(2)=(2-2)^{2}-2$

$g(2)=-2$

Then, the minimum point of this function is (2,-2) and it corresponds to (D)

I hope it helps you!

8 0

3 years ago

Solve the system of equations by substitution. 3/8 x + 1/3 y =17/24 and x + 7y = 8

goblinko [34]

$\left \{ {{ \frac{3}{8}x+ \frac{1}{3}y = \frac{17}{14} } \atop {x+7y=8}} \right.$

To solve this system by substitution, first isolate x in the second equation.

$x+7y=8$
$x+7y-7y=8-7y$
$x=8-7y$

Now, plug this expression ( $8-7y$ ) for x in the top equation to solve for y.

$\frac{3}{8} (8-7y)+ \frac{1}{3} y= \frac{17}{14}$
$3- \frac{21}{8} y+ \frac{1}{3} y= \frac{17}{24}$
$72-63y+8y=17$
$72-55y=17$
$-55y=17-72$
$-55y=-55$
$y=1$

Now that you have y, plug it into the second equation and solve for x.

$x+7y=8$
$x+7(1)=8$
$x+7=8$
$x=1$

Last step is to plug your x- and y-values in to both equations to check your work.

$\frac{3}{8} (1)+ \frac{1}{3} y= \frac{17}{24}$

$\frac{3}{8} * \frac{3}{3} = \frac{9}{24} ; \frac{1}{3} * \frac{8}{8} = \frac{8}{24}$

$\frac{9}{24} + \frac{8}{24} = \frac{17}{24}$ <--True

$1+7(1)=8$
$1+7=8$ <--True

Answer:
$x=1 \\ y=1$

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3 years ago

Simplify 8(xy + 8) + 1. 8xy + 9 8xy + 65 8xy + 73

mixer [17]

64x2y2 the twos are powers of ten so you put them up top.

8 0

4 years ago

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