All categories

3 years ago

If the measure of angle ACB is 50 degrees, what is the measure of angle AOB

Mathematics

1 answer:

kodGreya [7K]3 years ago

8 0

The answer is 100 degree

You might be interested in

Please answer this correctly

Bond [772]

Answer: The second option (purple model)

Step-by-step explanation: It is that option because there were 80 toys collected which is 100% and then we do not what the 30% is.

5 0

4 years ago

Read 2 more answers

The black graph is the graph of y=f(x). Choose the equation for the red graph.

Makovka662 [10]

Answer:

y-1 = f(x)

Step-by-step explanation:

Here, we want to choose the equation for the ref graph

The red graph as we can see is above the black

This means it is more positive

The difference between the two is just 1 unit

By the addition of 1 to the y-value of the black graph, we get the red

Thus, we have that;

y- 1 = f(x)

6 0

3 years ago

Here is my question

arsen [322]

Answer:

A. 2[24÷ (4+4)]

Step-by-step explanation:

Hope this helps :)

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E3_t%7B%7Cx%2B2%7C%7D%20%5C%2C%20dx%20" id="TexFormula1" title=" \int\limi

photoshop1234 [79]

$\displaystyle\int_{-3}^3|x+2|\,\mathrm dx$

Recall the definition of absolute value:

$|x|=\begin{cases}x&\text{for }x\ge0\\-x&\text{for }x$

So we can split up the integration interval at $x=-2$ and apply this definition to rewrite the integral as

$\displaystyle\int_{-3}^{-2}(-(x+2))\,\mathrm dx+\int_{-2}^3(x+2)\,\mathrm dx$
$=\displaystyle-\int_{-3}^{-2}(x+2)\,\mathrm dx+\int_{-2}^3(x+2)\,\mathrm dx$
$=-\left(\dfrac12x^2+2x\right)\bigg|_{x=-3}^{x=-2}+\left(\dfrac12x^2+2x\right)\bigg|_{x=-2}^{x=3}$
$=\dfrac12+\dfrac{25}2=13$

7 0

3 years ago

Find the zeros of the function and state their multiplicities. y=x^4-8x^2+16

Volgvan

$y=0\\\\ 
x^4-8x^2+16 = 0\\\\
(x^2)^2 - 2\cdot 4 \cdot x^2 + 4^2 = 0\\\\
(x^2 - 4)^2 = 0\\\\
(x^2 - 4)(x^2 - 4)=0\\\\
(x-2)(x+2) (x-2)(x+2) = 0\\\\
(x-2)(x-2) (x+2)(x+2) = 0\\\\
\Longrightarrow~~~ x_1 = 2;~x_2 = 2;~ x_3= -2;~x_4 = -2\\\\
\text{Conclusion:}\\\\
x_1 = x_2 = 2~~~ \text{double root}\\\\
x_3 = x_4 = -2~~~ \text{double root}\\\\
$

7 0

3 years ago