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3 years ago

1. Without using a calculator, find all the roots of the equation. *+x+4x+4 = 0

Mathematics

1 answer:

Stells [14]3 years ago

5 0

Answer:

Step-by-step explanation:

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The quotient of the square root of x and five

Tasya [4]

Answer:

$n \div 5 \sqrt{x}$

Step-by-step explanation:

Convert it to an expression:

$n \div 5 \sqrt{x}$

7 0

4 years ago

Read 2 more answers

Can someone please help me ☹️

Vadim26 [7]

Answer:

When c = 6 and d = 34, there is one solution. When c = 8 and d = 6, there is no solution. When c = 8 and d = 12, there are infinitely many solutions.

Step-by-step explanation:

6x + 34 = 8x +12

-6x -6x

34 = 2x +12

-12 -12

22 = 2x

11 = x

8x + 6 = 8x +12

-8x -8x

6 = 12

No solution

8x+12= 8x + 12

-8x-12 -8x - 12

0 = 0

Infinite solutions

7 0

3 years ago

The low temp for 5 days was -5 -7 -2 2 and -3 what was the average low temperature for those days

Arada [10]

Answer:

-3 degrees

Step-by-step explanation:

Find the average in the usual way: sum up the five temperatures and divide the result by 5:

-5 -7 -2 +2 -3 -15

-------------------- = ---------- = -3 degrees

5 5

7 0

4 years ago

Condense log₂ 4 + log₂ 5

yawa3891 [41]

Answer: $\log_2(20)$

Work Shown:

$\log_2(4) + \log_2(5)\\\\\log_2(4*5)\\\\\log_2(20)\\\\$

The rule used is $\log_{b}(\text{x}) + \log_{b}(\text{y}) = \log_{b}(\text{xy})$

7 0

2 years ago

Read 2 more answers

Find the side in of point a to point c

vladimir1956 [14]

Step 1:

Calculate the measure of angle ∠ABC

$\angle DBC+\angle ABC=180(\text{ sum of angles on a straight line)}$ $\angle ABC=65^0$ $\begin{gathered} \angle DBC+\angle ABC=180 \\ \angle DBC+65^0=180^0 \\ \angle DBC=180^0-65^0 \\ \angle DBC=115^0 \end{gathered}$

From the triangle in the question,

$a=10\operatorname{km},c=15\operatorname{km},B=115^0$

Step 2:

Calculate the value of AB using the cosine rule below

$b^2=a^2+c^2-2\times a\times c\times\cos B$

By substituting the values, we will have

$\begin{gathered} b^2=a^2+c^2-2\times a\times c\times\cos B \\ b^2=10^2+15^2-2\times10\times15\times\cos 115^0 \\ b^2=100+225-300\times(-0.4226) \\ b^2=325+126.78 \\ b^2=451.78 \\ \text{Square root both sides} \\ \sqrt[]{b^2}=\sqrt[]{451.78} \\ b=21.26\operatorname{km} \end{gathered}$

Hence,

The distance of point A to point C is = 21.26km

3 0

1 year ago