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3 years ago

The equation 9+3a-4-=?a+20-15 has infinitely many solutions. What value should be placed in the box?

Mathematics

1 answer:

Alina [70]3 years ago

7 0

Answer:

9 + 3a -4 - = a +20-15

3a - a = 20 -15+4-9

2a = 24 - 24

2a = 0

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A farm has a square fenced-in area with side length x where the farm gives children pony rides. Adjacent to one entire side of t

Vika [28.1K]

Answer:

1) x²+4x=2700

2) 50x50

Step-by-step explanation:

Pony area = x²

line-up area=4x

x²+4x=2700

x²+4x-2700=0

(x-50)(x+54)

So x= 50 or -54. Since it can’t be a negative number, it must be 50.

CHECK:

Square = 50x50 = 2500

Line up area = 50x4 = 200

2500+200=2700 YES

6 0

3 years ago

A trapezoid has bases of lengths 13 cm and 17 cm. The

Vedmedyk [2.9K]

Answer:

135 cm²

Step-by-step explanation:

The area (A) of a trapezoid is calculated as

A = $\frac{1}{2}$ h(a + b)

where h is the height and a, b the parallel bases

Here h = 9, a = 13 and b = 17 , thus

A = $\frac{1}{2}$ × 9 × (13 + 17)

= 0.5 × 9 × 30

= 135 cm²

3 0

3 years ago

For a Saturday matinee, adult tickets cost $8.50 and kids under 12 pay only $7.00. If 80 tickets are sold for a total

jolli1 [7]

Answer:

30 adult tickets,
50 kids tickets

Step-by-step explanation:

Given

Cost of adult ticket = $8.50
Cost of kids ticket = $7.00
Number of tickets = 80
Total cost = $605

Let the adult ticket be a and kids be k

We got equations

a + k = 80
8.5a + 7k = 605

From the first equation we get

k = 80 -a

Substituting k in the second equation

8.5a + 7(80 - a) = 605
8.5a - 7a + 560= 605
1.5a = 45
a = 45/1.5
a = 30

Then finding k

k = 80 - 30 = 50

8 0

3 years ago

Mrs. Carlyle bought a bag of nuts for her children. Phillip, Joy, Brent, and Preston took some from the bag.Phillip took 1/3. Jo

BARSIC [14]

Answer:

Originally there was 324 nuts in the bag.

Phillip took 108, Joy took 54, Brent took 81 and Preston took 10.

Step-by-step explanation:

Let's call the total amount of nuts N, and the number of nuts each child took by the initial of their name (Phillip will be P1 and Preston will be P2).

So we can write the following equations:

P1 = N/3

After removing N/3, the remaining nuts is (N - N/3):

J = (N - N/3)/4 = (2N/3)/4 = N/6

After removing N/6 from (N - N/3), the remaining nuts is (N - N/3 - N/6):

B = (N - N/3 - N/6)/2 = (N/2)/2 = N/4

P2 = 10

In the final there were 71 nuts remaining, so we have that:

N - N/3 - N/6 - N/4 - 10 = 71

N - N/3 - N/6 - N/4 = 81

N/4 = 81

N = 324 nuts

The amount of nuts took by each child is:

P1 = N/3 = 108 nuts

J = N/6 = 54 nuts

B = N/4 = 81 nuts

P2 = 10 nuts

8 0

3 years ago

In ΔOPQ, o = 9.2 cm, p = 2.4 cm and ∠Q=37°. Find the length of q, to the nearest 10th of a centimeter.

storchak [24]

The length of q, to the nearest 10th of a centimeter is 7.6 cm.

Given in question,

In ΔOPQ,

o = 9.2 cm

p = 2.4 cm

∠Q = 37°

Cosine formula ⇒ cos θ = $\frac{o^{2}+p^{2}-q^{2} }{2op}$

Putting the values in equation,

cos 37 = $\frac{(9.2)^{2}+(2.4)^{2}-q^{2} }{2*9.2*2.4}$

0.799 = $\frac{84.64 + 5.76-q^{2} }{44.16}$

0.799*44.16 = 90.4 - $q^{2}$

32.28 = 90.4 - $q^{2}$

$q^{2}$ = 90.4 - 32.28

$q^{2}$ = 58.12

$q = \sqrt{58.12}$

$q = 7.63$

q = 7.6 cm (to nearest 10th)

Hence, length of q is 7.6 cm.

Learn more about length on:

brainly.com/question/8552546

#SPJ1

3 0

1 year ago