If u have 3/4 cups of flour, how many 8th do u have

In the diagram ABC is congruent to DEF. aging the values of x and y

kipiarov [429]

Answer:

x = 1, y = 46

Step-by-step explanation:

Since the triangles are congruent then corresponding sides and angles are congruent, then

DE = AB , that is

3x + 6 = 9 ( subtract 6 from both sides )

3x = 3 ( divide both sides by 3 )

x = 1

and

∠ C = ∠ F

10x + y = 56 ( substitute x = 1 )

10(1) + y = 56

10 + y = 56 ( subtract 10 from both sides )

y = 46

5 0

3 years ago

Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x)= 6x^1/3 + 3x^4/3. You must justify

irina [24]

Answer:

x-coordinates of relative extrema = $\frac{-1}{2}$

x-coordinates of the inflexion points are 0, 1

Step-by-step explanation:

$f(x)=6x^{\frac{1}{3}}+3x^{\frac{4}{3}}$

Differentiate with respect to x

$f'(x)=6\left ( \frac{1}{3} \right )x^{\frac{-2}{3}}+3\left ( \frac{4}{3} \right )x^{\frac{1}{3}}=\frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}$

$f'(x)=0\Rightarrow \frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}=0\Rightarrow x=\frac{-1}{2}$

Differentiate f'(x) with respect to x

$f''(x)=2\left ( \frac{-2}{3} \right )x^{\frac{-5}{3}}+\frac{4}{3}x^{\frac{-2}{3}}=\frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}\\f''(x)=0\Rightarrow \frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}=0\Rightarrow x=1$

At x = $\frac{-1}{2}$ ,

$f''\left ( \frac{-1}{2} \right )=\frac{4\left ( -1+4\left ( \frac{-1}{2} \right ) \right )}{3\left ( \frac{-1}{2} \right )^{\frac{5}{3}}}>0$

We know that if $f''(a)>0$ then x = a is a point of minima.

So, $x=\frac{-1}{2}$ is a point of minima.

For inflexion points:

Inflexion points are the points at which f''(x) = 0 or f''(x) is not defined.

So, x-coordinates of the inflexion points are 0, 1

7 0

3 years ago

10-12 please help, find the missing number and once you found the missing number and answered it you make sure it’s the solution

irinina [24]

10. Infinitely many solutions:

$2(x-1)+6x=4(?-1)+2.$

Simplify this equation:

$2x-2+6x=4\cdot ?-4+2,\\ \\8x-2=4\cdot ? -2,\\ \\8x=4\cdot ?$

When $?=2x,$ the equation becomes

$8x=8x \text{ or } 0=0.$

This equality is true for all x, thus the equation has infinitely many solutions.

11. The equation $5x-(x-2)=2x-(?)$ should have one solution x=-1.

First, simplify the equation:

$5x-x+2=2x-?,\\ \\4x+2=2x-?,\\ \\2x+2=-?,\\ \\?=-2x-2.$

This equation will have solution x=-1 when

$?=-2\cdot (-1)-2=2-2=0.$

Therefore, the initial equation is

$5x-(x-2)=2x.$

12. The equation $-(x-8)+4x=2\cdot ?+x$ should have infinitely many solutions.

Simplify it:

$-x+8+4x=2\cdot ?+x,\\ \\2x+8=2\cdot ?,\\ \\2(x+4)=2\cdot ?.$

When $?=x+4,$ this equation will have infinitely many solutions.

6 0

3 years ago

Laura owes $4759 on a credit card with a 20.9% interest rate compounded monthly. What is the monthly payment she should make in

Juliette [100K]

This can be calculated using the formula:
P = L((r/n)*(1 + r/n)^(n*t))/((1 + r/n)^(n*t) - 1)
Where:
L = 4759
r = 0.209
n = 12
t = 3
So plugging in our data:P = 4759((0.209/12)*(1 + 0.209/12)^(12*3))/((1 + 0.209/12)^(12*3) - 1)
Which will give us the amount of: $179.05 is the monthly repayments.
Other info:
Total interest:$1,686.80
Total cost:$6,445.80

8 0

3 years ago

Read 2 more answers

Some of the students three scores is 231. If the first is 20 points more than the second, and the sum of the first two is 6 more

Aleks [24]

Answer:

The first score is 109

Step-by-step explanation:

I am assuming that in the first sentence of the question, you meant:

Sum of the students three scores is 231...

First, let the scores of the first second and third student be a, b and c respectively. We are told that:

a + b + c = 231 . . . . . . . .(1) (sum of students three scores is 231)

a = b + 20 . . . . . . . . . . . (2) (the first is 20 points more than the second)

a + b = 6c . . . . . . . . . . . .(3) (sum of the first two is 6 more times the third)

required, find a.

substituting the value of (a + b) in equation (3) into equation (1), we will have the following:

since a + b = 6c . . . (3)

a + b + c = 231 . . . . . (1), becomes,

(a + b) + c = 231

(6c) + c = 231

7c = 231 (divide both sides by 7)

c = 231 ÷ 7 = 33

∴ c = 33

Next, from equation (2), we know that a = b + 20; this can also be written as:

a - 20 = b

∴ b = a - 20 . . . . . . . (4)

Finally, putting the value of b in equation (4) and the value of c calculated above into equation 1, ( a + b + c = 231), we have the following:

a + (a - 20) + 33 = 231

(a + a) - 20 + 33 = 231

2a + 13 = 231

2a = 231 - 13 = 218

a = 218 ÷ 2 = 109

∴ a = 109

we can also calculate for 'b' by substituting for the value of 'a' in equation 4

b = a - 20 = 109 - 20 = 89.

and to test if the values of a, b and c are correct:

a + b + c = 231

109 + 89 + 33 = 231

4 0

3 years ago