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3 years ago

Which angles are supplementary to each other?

Mathematics

1 answer:

user100 [1]3 years ago

4 0

Answer:

Angles 2 and 3 are supplementary to each other

Step-by-step explanation:

A supplementary angle is 2 angles that when you add the sum is 180.

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10(5 - n) - 1 = 29<br><br> n = ?<br> Help!

nydimaria [60]

n = 2

distribute and simplify left side of equation

50 - 10n - 1 = 29

49 - 10n = 29 ( subtract 49 from both sides )

- 10n = 29 - 49 = - 20

n = $\frac{-20}{-10}$ = 2

5 0

3 years ago

Read 2 more answers

(x+3)^2=4

GalinKa [24]

Answer:

x = -1, -5

Step-by-step explanation:

${(x + 3)}^{2} = 4 \\ = > {x}^{2} + 2 \times x \times 3 + {3}^{2} = 4 \\ = > {x}^{2} + 6x + 9 = 4 \\ = > {x}^{2} + 6x = 4 - 9 \\ = > {x}^{2} + 6x = - 5 \\ = > {x}^{2} + 6x + 5 = 0 \\ = > {x}^{2} + 5x + x + 5 = 0 \\ = > x(x + 5) + 1(x + 5) = 0 \\ = > (x + 1)(x + 5) \\ = > x = - 1 \: \: or \: \: - 5$

Hope it helps.

Don't forget to click the "Brainliest" button if you like my answer.

6 0

2 years ago

Read 2 more answers

Let $f(x) = x^2$ and $g(x) = \sqrt{x}$. Find the area bounded by $f(x)$ and $g(x).$

Anna [14]

Answer:

$\large\boxed{1\dfrac{1}{3}\ u^2}$

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

$x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1$

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

$\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}$

6 0

2 years ago

A function models the growth of a sample starting with 40 bacteria with the ability to double every two hours. Which graph

natima [27]

Answer:

The function is y = 40 * 2^(x/2)

The graph is in the image attached

Step-by-step explanation:

The function that models this growth is an exponencial function, that can be described with the following equation:

y = a * b^(x/n)

Where a is the inicial value, b is the rate of growth, x is the time and n is the relation between the time and the rate (the rate occurs for every two hours, so n = 2).

Then, using a = 40, r = 2 and n = 2, we have:

y = 40 * 2^(x/2)

If we plot this function, we have the graph shown in the image attached,

It is an exponencial graph, where the value of y increases very fast in relation to the increase of x.

7 0

2 years ago

Mary Stahley invested $2500 in a 36-month certifi- cate of deposit (CD) that earned 8.5% annual simple interest. When the CD mat

nata0808 [166]

9514 1404 393

Answer:

$13,916.24

Step-by-step explanation:

First, we need to find the value of the CD at maturity.

A = P(1 +rt) . . . . simple interest rate r for t years

A = $2500(1 +0.085·3) = $2500×1.255 = $3137.50

Now, we can find the value of the account with compound interest.

A = P(1 +r)^t . . . . . rate r compounded annually for t years

A = $3137.50 × 1.18^9 = $13,916.24

The mutual fund was worth $13,916.24 after 9 years.

3 0

2 years ago