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3 years ago

Which of the following sets of numbers could not represent the three sides of a triangle?

Mathematics

1 answer:

vlabodo [156]3 years ago

8 0

Answer:

Step-by-step explanation:

The length of the longest member must be less than the sum of the two shorter members.

Cannot be triangle

(8, 10, 19) because 19 ≥ 10 + 8

The other three could be triangles

If the longest equals the sum of the two shorter, then the three line segments all lie along the same line and no triangle exists.

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Round to 2 SF 18.9 pls its urgent

lawyer [7]

Rounding 2 SF would give you 19

8 0

3 years ago

Which expressions are equivalent to (5g+3g+4)x2

FrozenT [24]

It’s A, cuz the 4*2 equals 8

4 0

3 years ago

What is the true solution to the equation below?

jarptica [38.1K]

Answer:

Step-by-step explanation:

some rules of logarithmic function

$ln(a) - ln(b)=ln(\frac{a}{b})$

$e^{ln(a)}=a$

$ln(a)^{n}=nln(a)$ vice-versa $nln(a)=ln(a)^{n}$

If ㏑(a) = ㏑(b), then a = b

∴ $2ln(e^{ln(2x)})-ln(e^{ln(10x)})=ln(30)$

Use the 2nd rule to simplify it

$e^{ln(2x)}=2x\\e^{ln(10x)}=10x\\$

2㏑(2x) - ㏑(10x) = ㏑(30)

Use the 3rd rule in the 1st term

∵ 2㏑(2x) = ㏑(2x)² = ㏑(4x²)

∴ ㏑(4x²) - ㏑(10x) = ㏑(30)

- Use the 1st rule with the left hand side

$ln(4x^{2})-ln(10x)=ln(\frac{4x^{2}}{10x})\\\\ln(\frac{4x^{2}}{10x})=ln(30)\\\\ \frac{4x^{2}}{10x}=\frac{2x}{5}=\frac{2}{5}x\\\\ ln(\frac{2}{5}x)=ln(30)$

Use the 4th rule

$\frac{2}{5} x = 30$

Multiply both sides by 5

∴ 2 x = 150

- Divide both sides by 2

∴ x = 75

The value of x = 75

3 0

3 years ago

HELP ME!! im so stumped. will give brainliest!

Olegator [25]

Answer:

$x=1+\sqrt{3}$ and $x=1-\sqrt{3}$

Step-by-step explanation:

We have been given the parabola with vertex (1, -9) and y intercept at (0, -6).

Now we need to find the x-intercepts of that parabola. So first we begin by finding the equation of parabola using vertex formula:

$y=a\left(x-h\right)^2+k$

Vertex for this formula is given by (h,k)

Compare that with given vertex (1,-9), we get: h=1, k=-9

So plug these into vertex formula:

$y=a\left(x-1\right)^2-9$ ...(i)

Plug given point (0, -6). into (i)

$-6=a\left(0-1\right)^2-9$

$-6=a\left(-1\right)^2-9$

$-6=a\left(1\right)-9$

$-6+9=a$

$3=a$

Plug a=3 into (i)

$y=3\left(x-1\right)^2-9$

Now to find x-intercept, we just plug y=0 and solve for x

$y=3\left(x-1\right)^2-9$

$0=3\left(x-1\right)^2-9$

$9=3\left(x-1\right)^2$

$3=\left(x-1\right)^2$

$\pm \sqrt{3}=x-1$

$1+\pm \sqrt{3}=x$

Hence final answer are

$x=1+\sqrt{3}$ and $x=1-\sqrt{3}$

4 0

3 years ago

Draw the line through the point (4,5) that is parallel to the y-axis. Write the equation of this line and state its slope

ollegr [7]

Parallel to the y axis...that would mean that it is a vertical line.
The equation would be : x = 4 and it would have an undefined slope

7 0

3 years ago