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3 years ago

Let's Check In

Mathematics

1 answer:

crimeas [40]3 years ago

8 0

Answer:

I need more info about the shape were making but I think C is the answer.

Step-by-step explanation:

I'll be honest Im confused. Im going to assume your finding the third side of triangle and if so here is my answer.

Side GH must be between 3.5 and 23.1 not inclusive because 13.3+9.8= 23.1 and that would be the length if the line is a straight line, and the minimum is 13.3-9.8= 3.5 and that is only if the lines overlap. That only leaves C as the answer within the range.

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Can someone plz help me with this

slava [35]

Answer:

Step-by-step explanation:

2a. f(-3) = (-3)^2-4(-3)+6 = 9+12+6 = 27

2b. f(-2) = 3(-2)^2-(-2)^3 -2 +1 = 3(4) - (-8) -2 +1

= 12+8-1 = 19

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2 years ago

Express the form 1: n <br><br> 5:20

Stella [2.4K]

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5 : 20

Divide both sides by 5 to get the answer

5/5 = 1

20/5 = 4

Thus, the answer is 1 : 4

Have A Nice Day ❤

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3 years ago

(NEED HELP ASAP!!!) Jamie unfolded a cardboard box. The figure of the unfolded box is shown below: A long rectangle divided alon

Oksanka [162]

Answer:

6 x 4 x 4 Hope I get u right<3

Step-by-step explanation:

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3 years ago

Find the exact value of sec30°.

julsineya [31]

Decimal Form:

1.15470053

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Step-by-step explanation:

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3 years ago

Read 2 more answers

A donut store has 11 different types of donuts. You can only buy a bag of 3 of them, where each donut has to be of a different t

MakcuM [25]

Answer:

$165$ .

Step-by-step explanation:

Since repetition isn't allowed, there would be $11$ choices for the first donut, $(11 - 1) = 10$ choices for the second donut, and $(11 - 2) = 9$ choices for the third donut. If the order in which donuts are placed in the bag matters, there would be $11 \times 10 \times 9$ unique ways to choose a bag of these donuts.

In practice, donuts in the bag are mixed, and the ordering of donuts doesn't matter. The same way of counting would then count every possible mix of three donuts type $3 \times 2 \times 1 = 6$ times.

For example, if a bag includes donut of type $x$ , $y$ , and $z$ , the count $11 \times 10 \times 9$ would include the following $3 \times 2 \times 1$ arrangements:

$xyz$ .
$xzy$ .
$yxz$ .
$yzx$ .
$zxy$ .
$zyx$ .

Thus, when the order of donuts in the bag doesn't matter, it would be necessary to divide the count $11 \times 10 \times 9$ by $3 \times 2 \times 1 = 6$ to find the actual number of donut combinations:

$\begin{aligned} \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

Using combinatorics notations, the answer to this question is the same as the number of ways to choose an unordered set of $3$ objects from a set of $11$ distinct objects:

$\begin{aligned}\begin{pmatrix}11 \\ 3\end{pmatrix} &= \frac{11 !}{(11 - 3)! \times 3 !} \\ &= \frac{11 !}{8 ! \times 3 !} \\ &= \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

5 0

2 years ago