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Julli [10]
2 years ago
5

Please answer this question I need to answer please answer HURRY!!!!

Mathematics
2 answers:
Goryan [66]2 years ago
6 0

Answer:

52

Step-by-step explanation:

r^{2} -14+8p

Substitute values given:

(8)^{2} -14+8(\frac{1}{4} )

Simplify:

64-14+2

50+2

52

Therefore 52 is our answer

VikaD [51]2 years ago
5 0

Answer:

56

Step-by-step explanation:

if r = 8 than r to the second power is 8 to the second power (or 8x8) which equals 64

64 - 14 + 8p

If p + 1/4 = 2

64 - 14 + 2

54 + 2

56

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gladu [14]

Answer:

b. Multiply 15 by 9, then divide by 2

Step-by-step explanation:

15 / 2/9 = 15 * 9/2

Answer: b. Multiply 15 by 9, then divide by 2

6 0
3 years ago
The picture this is math 1 for apex learning
noname [10]

Answer:

C.

Step-by-step explanation:

(f+g)(x) ➡ f(x) + g(x)

x/2 - 2 + 2x^2 + x - 3 = 2x^2 + 3x/2 -5

3 0
3 years ago
Pls help me I don't understand​
frutty [35]

Answer:

d. \:  \:  \: 25 {x}^{6}

Step-by-step explanation:

Given,

Side of the square

= 5 {x}^{3}

Therefore, area of the square = side × side

=  {5x}^{3}  \times 5 {x}^{3}

= 5 \times 5 \times  {x}^{3 + 3}

= 25 {x}^{6}

8 0
2 years ago
Read 2 more answers
Help me please. <br>Help me please. <br>Help me please.
Gnesinka [82]
A)
To be similar triangles have to have equal angles
triangle ZDB' 
1)angle Z=90 degrees
triangle B'CQ
1) angle C 90 degrees

angle A'B'Q=90
DB'Z+A'B'Q+CB'Q=180, straight angle
DB'Z+90+CB'Q=180
DB'Z+CB'Q=90

triangle ZDB'
DZB'+DB'Z=180-90=90

DB'Z+CB'Q=90
DZB'+DB'Z=90
DB'Z+CB'Q=DZB'+DB'Z
2)CB'Q=DZB' (these angles from two triangles ZDB' and B'CQ )
3)so,angles DB'Z and B'QC are going to be equal because of sum of three angles in triangles =180 degrees and 2 angles already equal.
so this triangles are similar by tree angles
b)
B'C:B'D=3:4
B'D:DZ=3:2
CQ-?
DC=AB=21
DC=B'C+B'D (3+4= 7 parts)
21/7=3
B'C=3*3=9
B'D=3*4=12
B'D:DZ=3:2
12:DZ=3:2
DZ=12*2/3=8
B'D:DZ=CQ:B'C
3:2=CQ:9
CQ=3*9/2=27/2

c)
BC=BQ+QC=B'Q+QC
BQ' can be found by pythagorean theorem



3 0
3 years ago
Help me on these questions
mojhsa [17]

Answer:

a) The equation is (y - 1)² = -8 (x - 4)

b) The equation is (x - 1)²/25 + (y - 4)²/16 = 1

c) The equation of the ellipse is (x - 3)²/16 + y²/4 = 1

Step-by-step explanation:

a) Lets revise the standard form of the equation of the parabola with a

   horizontal axis

# (y - k)² = 4p (x - h), (h , k) are the coordinates of its vertex and p ≠ 0

- The focus of it is (h + p , k)

* Lets solve the problem

∵ The focus is (2 , 1)

∵ focus is (h + p , k)

∴ h + p = 2 ⇒ subtract p from both sides

∴ h = 2 - p ⇒ (1)

∴ k = 1

∵ It opens left, then the axis is horizontal and p is negative

∴ Its equation is (y - k)² = 4p (x - h)

∵ k = 1

∴ Its equation is (y - 1)² = 4p (x - h)

- The parabola contains point (2 , 5), substitute the coordinates of the

 point in the equation of the parabola

∴ (5 - 1)² = 4p (2 - h)

∴ (4)² = 4p (2 - h)

∴ 16 = 4p (2 - h) ⇒ divide both sides by 4

∴ 4 = p (2 - h) ⇒ (2)

- Use equation (1) to substitute h in equation (2)

∴ 4 = p (2 - [2 - p]) ⇒ open the inside bracket

∴ 4 = p (2 - 2 + p) ⇒ simplify

∴ 4 = p (p)

∴ 4 = p² ⇒ take √ for both sides

∴ p = ± 2, we will chose p = -2 because the parabola opens left

- Substitute the value of p in (1) to find h

∵ h = 2 - p

∵ p = -2

∴ h = 2 - (-2) = 2 + 2 = 4

∴ The equation of the parabola in standard form is

  (y - 1)² = 4(-2) (x - 4)

∴ The equation is (y - 1)² = -8 (x - 4)

b) Lets revise the equation of the ellipse

- The standard form of the equation of an ellipse with  center (h , k)

 and major axis parallel to x-axis is (x - h)²/a² + (y - k)²/b² = 1  

- The coordinates of the vertices are (h ± a , k )  

- The coordinates of the foci are (h ± c , k), where c² = a² - b²  

* Now lets solve the problem

∵ Its vertices are (-4 , 4) and (6 , 4)

∵ The coordinates of the vertices are (h + a , k ) and (h - a , k)  

∴ k = 4

∴ h + a = 6 ⇒ (1)

∴ h - a = -4 ⇒ (2)

- Add (1) and (2) to find h

∴ 2h = 2 ⇒ divide both sides by 2

∴ h = 1

- Substitute the value of h in (1) or (2) to find a

∴ 1 + a = 6 ⇒subtract 1 from both sides

∴ a = 5

∵ The foci at (-2 , 4) and (4 , 4)

∵ The coordinates of the foci are (h + c , k) , (h - c , k)

∴ h + c = 4

∵ h = 1

∴ 1 + c = 4 ⇒ subtract 1 from both sides

∴ c = 3

∵ c² = a² - b²

∴ 3² = 5² - b²

∴ 9 = 25 - b² ⇒ subtract 25 from both sides

∴ -16 = -b² ⇒ multiply both sides by -1

∴ 16 = b²

∵ a² = 25

∵ The equation of the ellipse is (x - h)²/a² + (y - k)²/b² = 1

∴ The equation is (x - 1)²/25 + (y - 4)²/16 = 1

c) How to identify the type of the conic  

- Rewrite the equation in the general form,  

 Ax² + Bxy + Cy² + Dx + Ey + F = 0  

- Identify the values of A and C from the general form.  

- If A and C are nonzero, have the same sign, and are not equal  

 to each other, then the graph is an ellipse.  

- If A and C are equal and nonzero and have the same sign, then

 the graph is a circle  

- If A and C are nonzero and have opposite signs, and are not equal  

 then the graph is a hyperbola.  

- If either A or C is zero, then the graph is a parabola  

* Now lets solve the problem

∵ x² + 4y² - 6x - 7 = 0

∵ The general form of the conic equation is

   Ax² + Bxy + Cy² + Dx + Ey + F = 0  

∴ A = 1 and C = 4

∵ If A and C are nonzero, have the same sign, and are not equal  to

  each other, then the graph is an ellipse.

∵ x² + 4y² - 6x - 7 = 0 ⇒ re-arrange the terms

∴ (x² - 6x ) + 4y² - 7 = 0

- Lets make x² - 6x completing square

∵ 6x ÷ 2 = 3x

∵ 3x = x × 3

- Lets add and subtract 9 to x² - 6x to make the completing square

 x² - 6x + 9 = (x - 3)²

∴ (x² - 6x + 9) - 9 + 4y² - 7 = 0 ⇒ simplify

∴ (x - 3)² + 4y² - 16 = 0 ⇒ add 16 to both sides

∴ (x - 3)² + 4y² = 16 ⇒ divide all terms by 16

∴ (x - 3)²/16 + 4y²/16 = 1 ⇒ simplify

∴ (x - 3)²/16 + y²/4 = 1

∴ The equation of the ellipse is (x - 3)²/16 + y²/4 = 1

5 0
3 years ago
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