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Dafna11 [192]
2 years ago
10

In London today, four times the high temperature was more than twice the high temperature plus

Mathematics
1 answer:
4vir4ik [10]2 years ago
4 0

Answer:

Let's define the high temperature as T.

We know that:

"four times T, was more than 2*T plus 66°C"

(i assume that the temperature is in °C)

We can write this inequality as:

4*T > 2*T + 66°C

Now we just need to solve this for T.

subtracting 2*T in both sides, we get:

4*T - 2*T > 2*T + 66°C - 2*T

2*T > 66°C

Now we can divide both sides by 2:

2*T/2 > 66°C/2

T > 33°C

So T was larger than 33°C

Notice that T = 33°C is not a solution of the inequality, then we should use the symbol ( for the set notation.

Then the range of possible temperatures is:

(33°C, ...)

Where we do not have an upper limit, so we could write this as:

(33°C, ∞°C)

(ignoring the fact that ∞°C is something impossible because it means infinite energy, but for the given problem it works)

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Leno4ka [110]

Answer:

greater

Step-by-step explanation: 45 and 40

8 0
3 years ago
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Find the value of the discriminant for <img src="https://tex.z-dn.net/?f=7x%5E%7B2%7D%20%2B5x%2B1%3D0" id="TexFormula1" title="7
kondor19780726 [428]

Answer:

No real roots

Step-by-step explanation:

Given

7x² + 5x + 1 = 0 ← in standard form

with a = 7, b = 5, c = 1

To determine the nature of the roots use the discriminant

Δ = b² - 4ac

• If b² - 4ac > 0 then roots are real and distinct

• If b² - 4ac = 0 then roots are real and equal

• If b² - 4ac < 0 then the roots are not real

Here

b² - 4ac = 5² - (4 × 7 × 1) = 25 - 28 = - 3

Thus the 2 roots are not real

7 0
3 years ago
A board 8 ft long is cut into three equals pieces. How long are the pieces?
san4es73 [151]

Answer:

A

Step-by-step explanation:

5 0
3 years ago
How to write g(x)=2x+3/x+1 in the form of:g(x)=a/x+p +q
aliya0001 [1]
Hello,

Please write correctly.

g(x)= (2x+3)/(x+1) to be written like a/(x+p) + q

It is very simple!

(2x+3)/(x+1)= [(2x+2)+1]/(x+1)= (2(x+1)+1)/(x+1)= 2+1/(x+1)


7 0
3 years ago
Need help now!! 20 points!!
Gelneren [198K]

When you have something like this, all you need to do is substitute the values, the last is for what value of x

For the first one;

((x^2+1)+(x-2))(2)

(x^2+x-1)(2)

(2)^2+(2)-1

4+2-1

5

For the second one;

((x^2+1)-(x-2))(3)

(x^2-x+3)(3)

(3)^2-(3)+3

9-3+3

9

For the last one;

3(x^2+1)(7)+2(x-2)(3)

3((7)^2+7)+2((3)-2)

3(49+7)+2(3-2)

3(56)+2(1)

168+2

170

3 0
2 years ago
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