Ask question

Ask question

All categories

2 years ago

5

6. The height of a triangle is 2 units more than the base. The area of the triangle is 10 square units. Find

1 answer:

Harrizon [31]2 years ago

7 0

Answer:

4.24

Step-by-step explanation:

First, let's set up the equation: $\frac{2+x^2}{2}=10$

Next, clear the fraction by multiplying both sides by 2: $2*(\frac{2+x^2}{2}=10)=2+x^2=20$

Then, subtract 2 from both sides: $x^2=18$

Finally, take a square root of both sides and round: $\sqrt{x^2=18} =$ (Rounded to the nearest hundredth) 4.24

You might be interested in

Divide recently purchased a pair of running shoes he got a 20% discount which save them nine dollars what was the original price

andrezito [222]

The original price of the running shoes was $45. Divide got a 20% discount on them and saved $9, and had to pay only pay $36.

- Lyla

7 0

3 years ago

Choose the fragment that best completes the following sentence. If you have a high credit score, you are more likely to

Oksi-84 [34.3K]

C, pay your balance in full

6 0

3 years ago

Read 2 more answers

What are the missing parts that correctly complete the proof? Given: Point A is on the perpendicular bisector of segment P Q. Pr

nikdorinn [45]

Answer:

I just finished the test with an 82. Here are the answers

I hope this helps and brainliest would be greatly appreciated

6 0

3 years ago

1.- What is 'a' for this hyperbola?

SpyIntel [72]

Answer:

The standard form of a hyperbola with vertices and foci on the x-axis:

$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$

where:

center: (h, k)
vertices: (h+a, k) and (h-a,k)
Foci: (h+c, k) and (h-c, k) where the value of c is c² = a² + b²
Slopes of asymptotes: $\pm\left(\dfrac{b}{a}\right)$

<h3><u>Part 1</u></h3>

The center of the given hyperbola is (0, 0), therefore:

$\implies \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$

Therefore $(\pm a,0)$ are the vertices. From inspection of the graph, $a=2$ .

<h3><u>Part 2</u></h3>

Choose two points on the asymptote with the positive slope:

(0, 0) and (4, 6)

Use the slope formula to find the slope:

$\sf slope\:(m)=\dfrac{change\:in\:y}{change\:in\:x}=\dfrac{6-0}{4-0}=\dfrac{3}{2}$

<h3><u>Part 3</u></h3>

Use the <u>slopes of asymptotes</u> formula, compare with the slope found in part 2:

$\implies \dfrac{b}{a}=\dfrac{3}{2}$

Therefore, $b=3$

<h3><u>Part 4</u></h3>

Substitute the found values of $a$ and $b$ into the equation from part 1:

$\implies \dfrac{x^2}{2^2}-\dfrac{y^2}{3^2}=1$

$\implies \dfrac{x^2}{4}-\dfrac{y^2}{9}=1$

4 0

2 years ago

Write the expression as a single natural logarithm. 3 In m + 4 Inn

skad [1K]

Answer:

$ln( {m}^{3} {n}^{4} )$

Step-by-step explanation:

$\alpha ln(x) = ln( {x}^{ \alpha } )$

$ln( {m}^{3} ) + ln( {n}^{4} )$

$ln(a) + ln(b) = ln(ab)$

$ln( {m}^{3} {n}^{4} )$

8 0

2 years ago

Read 2 more answers