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mafiozo [28]
3 years ago
6

346772.99378 x 222222666886=

Mathematics
1 answer:
nordsb [41]3 years ago
6 0

Answer:

7.706081948 lowecase10uppercase16

Step-by-step explanation:

I hope it helps u.

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(3-2sqrt-18)-(2+3sqrt-8)​
Mama L [17]
Tell me the question and I’ll help you with it is it simplify or evalute etc.
7 0
3 years ago
Pls help (kinda easy)
3241004551 [841]

Answer:

f(x)=(2x+1)(2x-1).

Step-by-step explanation:

We want to convert the function into the form that let's us easily find the x-intercept, and it would be for the form (ax+b)(cx+d) because then we can find the x-intercept in the following manner:

(ax+b)(cx+d)=0

x=-b/a

x=-d/c

We factor our function f(x)=4x^2-1 and get

\boxed{f(x)=(2x+1)(2x-1)}

Now this form let's us easily find the x-intercepts:

x=-1/2

x=1/2

and so we pick the second choice: f(x)=(2x+1)(2x-1).

5 0
3 years ago
Read 2 more answers
Quadrilateral jklm is a rhombus. the diagonals intersect at n. if the measure of angle jkl is 104°, find the measure of angle jk
Paul [167]
The diagram of rhombus JKLM is shown in the diagram below

A rhombus is a quadrilateral with four equal sides and its diagonals intersect perpendicular to each other (makes 90° angles). Opposite angles are equal (the same with a parallelogram). Each diagonal bisects the angle at J, K, L, and M equally

If angle JKL is 104°, the measurement of angle JKN is 104÷2=51°

5 0
2 years ago
Read 2 more answers
Find the measure of angle y. Round your answer to the nearest hundredth. (please type the numerical answer only)
Eduardwww [97]

Given:

The figure of a right triangle. PO=12,m\angle O=y^\circ and perpendicular = 12.

To find:

The measure of angle y.

Solution:

In a right angle triangle,

\sin\theta=\dfrac{Perpendicular}{Hypotenuse}

In the given right angle triangle,

\sin y^\circ=\dfrac{7}{12}

y^\circ=\sin ^{-1}\dfrac{7}{12}

y^\circ=35.685335^\circ

y^\circ\approx 35.69^\circ

Therefore, the measure of angle y is y^\circ =35.69^\circ.

7 0
3 years ago
The volume of a rectangular box with a square base remains constant at 500 cm3 as the area of the base increases at a rate of 6
Andre45 [30]

Answer:

the rate at which the height of the box is decreasing is -0.0593 cm/s

Step-by-step explanation:

Given the data in the question;

Constant Volume of a rectangular box with a square base = 500 cm³

area of the base increases at a rate of 6 cm²/sec

so change in the area of the base with respect to time dA/dt = 6 cm²/sec

each side of the base is 15 cm long

so Area of the base = 15 cm × 15 cm = 225 cm²

the rate at which the height of the box is decreasing = ?

Now,

V = Ah

dv/dt = 0 ⇒ Adh/dt + hdA/DT = 0

⇒ dh/dt = -hdA/dt / A

we substitute

dh/dt = [ -( 500 / 225 ) × 6 ] / 225

dh/dt = [ -(2.22222 × 6)  ] / 225

dh/dt = [ -13.3333 ] / 225

dh/dt = -0.0593 cm/s

Therefore, the rate at which the height of the box is decreasing is -0.0593 cm/s

7 0
2 years ago
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