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3 years ago

Assume EFG JKL. If m J = 90 , m K = 26 , and m L = 64 , what is the measure of E?

Mathematics

2 answers:

tatyana61 [14]3 years ago

7 0

Answer:

I think its 90

Step-by-step explanation:

Vlad [161]3 years ago

4 0

Answer:

E=90

Step-by-step explanation:

because EFG=JKL E=J F=K G=L

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NOT GRADED. FIND THE MEASURE OF THE INDICATER ANGLE IN EACH TRIANGLE.

Dennis_Churaev [7]

If it’s not graded does it really matter but it’s 4042 I think

3 0

3 years ago

The area of a circle is 43.25cm2. Find the length of the radius rounded to 2 DP.

Monica [59]

Answer:

3.71 cm

Step-by-step explanation:

Hi there!

Area of a circle equation: $A=\pi r^2$ where r is the radius

Plug in the area 43.25 cm²

$43.25=\pi r^2$

Divide both sides by π to isolate r²

$\frac{43.25}{\pi} =\frac{\pi r^2}{\pi} \\\frac{43.25}{\pi}=r^2$

Take the square root of both sides to isolate r

$\sqrt{\frac{43.25}{\pi}} =r\\3.71=r$

Therefore, the length of the radius rounded to 2 decimal points is 3.71 cm.

I hope this helps!

8 0

3 years ago

Find the absolute maximum and minimum values of the function below. f(x) = x3 − 9x2 + 3, − 3 2 ≤ x ≤ 12 Solution Since f is cont

Neko [114]

Answer:

There are an absolute minimum (x = 6) and an absolute maximum (x = 12).

Step-by-step explanation:

The correct statement is described below:

Find the absolute maximum and minimum values of the function below:

$f(x) = x^{3}-9\cdot x^{2}+ 3$ , $2 \leq x \leq 12$

Given that function is a polynomial, then we have the guarantee that function is continuous and differentiable and we can use First and Second Derivative Tests.

First, we obtain the first derivative of the function and equalize it to zero:

$f'(x) = 3\cdot x^{2}-18\cdot x$

$3\cdot x^{2}-18\cdot x = 0$

$3\cdot x \cdot (x-6) = 0$ (Eq. 1)

As we can see, only a solution is a valid critical value. That is: $x = 6$

Second, we determine the second derivative formula and evaluate it at the only critical point:

$f''(x) = 6\cdot x -18$ (Eq. 2)

x = 6

$f''(6) = 6\cdot (6)-18$

$f''(6) =18$ (Absolute minimum)

Third, we evaluate the function at each extreme of the given interval and the critical point as well:

x = 2

$f(2) = 2^{3}-9\cdot (2)^{2}+3$

$f(2) = -25$

x = 6

$f(6) = 6^{3}-9\cdot (6)^{2}+3$

$f(6) = -105$

x = 12

$f(12) = 12^{3}-9\cdot (12)^{2}+3$

$f(12) = 435$

There are an absolute minimum (x = 6) and an absolute maximum (x = 12).

6 0

3 years ago

can you help me please it's due in an hour and it's basically saying to find the slope please ignore the doodle

Anit [1.1K]

Answer:

$\frac{10}{-1}$

Step-by-step explanation:

slope = $\frac{rise}{run}$

$\frac{y_{2 - y_{1} } }{x_{2} - x_{3} }$ = $\frac{60 - 30}{3 - 6}$ = $\frac{30}{-3}$

simplify:

$\frac{30}{-3} = \frac{10}{-1}$

5 0

3 years ago

The sum of two consecutive odd integers is 28. What are the integers?

Fantom [35]

The correct option is B): "Let x = 1st integer. Let x + 2 = 2nd integer."
Given that "The sum of two consecutive odd integers is 28."

Now the difference between two odd numbers is 2. So this means that one odd number has to be greater than the other by 2.

Let the smaller number be 'X' therefore the bigger number will be 'x+2' thus x + (x+2) = 28

therefore, x + x + 2 = 28 now we subtract "2" from both the sides
therefore, 2x = 26 now dividing throughout by 2 we get:

x = 13 therefore, x + 2 = 13+2
= 15

therefore, the numbers are 13 AND 15

4 0

4 years ago