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quester [9]
3 years ago
14

Can u pls help me with this question ​

Mathematics
2 answers:
luda_lava [24]3 years ago
8 0
The answer is $15.58
mote1985 [20]3 years ago
8 0

Answer:

15.58

Step-by-step explanation:

it is

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Use completing the square to solve (x – 12)(x + 4) = 9 for x. x = –1 or 15 x = 1 or 7
4vir4ik [10]
(x - 12)(x + 4) = 9
x² + 4x - 12x - 48 = 9
x² - 8x - 48 = 9    |add 48 to both sides
x² - 8x = 57
x² - 2x · 4 = 57    |add 4² to both sides
x² - 2x · 4 + 4² = 57 + 4²    |use (a - b)² = a² - 2ab + b²
(x - 4)² = 57 + 16
(x - 4)² = 73 ⇔ x - 4 = √73 or x - 4 = -√73    |add 4 to both sides

x = 4 + √73 or x = 4 - √73
6 0
3 years ago
Read 2 more answers
If one card is drawn from a deck, find the probability of getting these results.
RSB [31]

Answer:

Face card= 12/52

(52 cards in a deck and 12 are face cards)

Red face card= 6/12

(12 face cards in a deck cards in a deck and 6 are red face cards)

Black face card= 6/52

(6 are black)

Black card= 26/52

(52 cards in a deck and 26 are black)

Red card= 26/52

(26 are red)

3 0
3 years ago
What shapes are formed by cross sections parallel to the base of a hexagonal pyramid?
siniylev [52]

The answer is the option B, which is: B. Similar hexagons.

The explanation for this answer is shown below:

By definition, a cross section is the shape that is formed when a plane cuts or makes a slice through a solid.

The hexagonal pyramid has an hexagonal as its base. If you make slices parallel to the base of this pyramid, you will see the shape of this base in different sizes. Therefore the shapes that are formed by cross sections parallel to the base of the hexagonal pyramid are similar hexagons.

7 0
3 years ago
Please explain step-by-step. If you don't explain, don't even bother answering.
Andre45 [30]

Answer:

<u>45°</u>

Step-by-step explanation:

From the diagram :

  • ∠AOC = ∠AOB + ∠BOC
  • 96 = 9x - 75 + 8x - 67
  • 17x - 142 = 96
  • 17x = 238
  • x = 238/17
  • x = 14

Finding ∠BOC :

  • ∠BOC = 8x - 67
  • ∠BOC = 8(14) - 67
  • ∠BOC = 112 - 67
  • ∠BOC = <u>45°</u>
4 0
2 years ago
Read 2 more answers
Given: KPST is a trapezoid, KP=ST, MN is a midsegment, MN=20, h=15, PS:KT=3:7 Find: KS and KP
lions [1.4K]

Answer:

Length of KS = 25 units and

Length of KP = 17 units.

Step-by-step explanation:

Given: KPST is a trapezoid, KP =ST, MN is a mid segment, h is the height =15

Also it is given that, MN = 20 , h = 15 and PS:KT = 3:7.

  • If two sides of the trapezoid are equal then, it is an isosceles trapezoid.
  • Mid-segment of a trapezoid is a line segment which connects the midpoints of the non-parallel sides.
  • A trapezoid mid segment connects the midpoints of two congruent sides of the trapezoid and is parallel to the pair of parallel sides.
  • Length of the mid segment is the sum of two bases divide by 2

Since, the length of Mid-segment MN = 20.

Also, it is given: PS:KT = 3:7

Let  PS = 3x and KT = 7x respectively.

then;

\frac{PS+KT}{2} = 20

\frac{3x+7x}{2} = 20

\frac{10x}{2} =20

On Simplify, we get;

5x =20

Divide both sides by 5 we get;

x =4

Then:

Length of base PS = 3x = 3(4) = 12 and Length of base KT = 7x = 7(4) = 28.

Now, In triangle PLK

Using Pythagoras theorem to find KP;

It is given here, PL =h =15 and KL= 8 {you can see in the figure as shown below};

KP^2= PL^2 + KL^2

KP^2 = 15^2+8^2 =225 +64 = 289

KP = \sqrt{289}

Simplify:

KP = 17

therefore, the length KP = 17 units

To, construct a line: Join K and S

Now, in triangle KRS

KR = KL +LR = 8 +12 =20 and SR = h= 15

Using Pythagoras theorem in KRS to find KS;

KS^2 = SR^2+KR^2

KS^2 = 15^2 + 20^2 = 225 + 400 = 625

KS = \sqrt{625}

On simplify:

KS = 25

Therefore, the length of KS is, 25 units.

4 0
3 years ago
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