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Delicious77 [7]
3 years ago
7

(b) At noon on Saturday, the temperature was 15 °C.

Mathematics
1 answer:
Thepotemich [5.8K]3 years ago
5 0
Their is a 18 degree difference
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The price of a ring was decreased by 25% to £390. what was the price before the decrease?
sergiy2304 [10]

Answer:

The original price was 520

Step-by-step explanation:

To find this, we first need to note that we paid 75% of the price. This is because we took 25% off from the original. Now we take the price we paid and divide it by the percentage of it which we paid. This will give us the original price.

390/75% = Total

390/.75 = Total

520 = Total

3 0
2 years ago
What is the simplified form of the quantity of x plus 8, all over the quantity of 3x plus 7 + the quantity of x plus 7, all over
Temka [501]
\frac{x+8}{3x+7} + \frac{x+7}{x+4}  \\ = \frac{(x+8)(x+4)+(x+7)(3x+7)}{(3x+7)(x+4)}  \\ = \frac{ x^{2} +12x+32+3 x^{2} +28x+49}{3 x^{2} +19x+28}  \\ = \frac{4 x^{2} +40x+81}{3 x^{2} +19x+28}
6 0
3 years ago
Read 2 more answers
Find the product or quotient for each. simple form<br> 5/9 divide 5/6
MakcuM [25]

Answer: 2/3

Step-by-step explanation:

5/9 and 5/6 doesn't have the same denominator. So 5/9 changes to 10/18, while 5/6 changes to 15/18.

10/18 divided by 15/18= 10/18 x 18/15 = 180/270

180/270= 2/3

8 0
3 years ago
How do I simplify the expression 7^0+7^2 over (7^3)^2
AlekseyPX

\frac{7^0+7^2}{(7^3)^2}=\frac{1+49}{7^6} = \frac{50}{117649}

Ok done. Thank to me :>

7 0
2 years ago
A ball is thrown vertically upward from the top of a building 160 feet tall with an initial velocity of 48 feet per second. The
alina1380 [7]

Answer:

# after 5 seconds, the ball strikes the ground

# The ball reaches maximum height at t = 1.5 seconds

# The max height is 196 feet

Step-by-step explanation:

The equation is:

d(t)=-16t^2+48t+160

t is the time

d(t) is the distance traveled

Initial height is 160 feet and initial velocity is 48 ft/sec

If we want to find the time it takes the ball to hit the ground, we let d(t) equal to 0 and find t. Shown below:

-16t^2+48t+160=0\\t^2-3t-10=0\\(t-5)(t+2)=0\\t=5,-2

We disregard t = -2 since time can't be negative. We take t = 5

Thus, after 5 seconds, the ball strikes the ground.

The equation is a quadratic of the form: ax^2+bx+c

Matching equations, we can say:

a = -16

b = 48

c = 160

The time when ball reaches max height is given as:

t=-\frac{b}{2a}

Substituting, we find:

t=-\frac{b}{2a}\\t=-\frac{48}{2(-16)}\\t=1.5

The ball reaches maximum height at t = 1.5 seconds

The max height can be found by putting t = 1.5 into the original equation. Shown below:

d(t)=-16t^2+48t+160\\d(1.5)=-16(1.5)^2+48(1.5)+160\\=196

The max height is 196 feet

8 0
3 years ago
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