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tino4ka555 [31]
2 years ago
5

The owner of Livingston Florist is assembling flower arrangements for Valentine's Day. This morning, she assembled 9 small arran

gements and 3 large arrangements, which took her a total of 78 minutes. After lunch, she arranged 1 large arrangement, which took 8 minutes. How long does it take to assemble each type?
Mathematics
1 answer:
Dmitriy789 [7]2 years ago
3 0

Answer:

24 minutes for 3 large arrangements

54 minutes for 9 small arrangements

6 minutes for each small arrangement

8 minutes for each large arrangement  

Step-by-step explanation:

hope this helps!

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Please help me ! thank you
denis-greek [22]

Answer:

57

Step-by-step explanation:

3 * 3 * 3 = 27 +       5*6 =30   27+30=57

Hope this helps!

8 0
2 years ago
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A bicyclist is riding on a path modeled by the function f(x) = 0.03(10x − x2), where x and f(x) are measured in miles. Find the
REY [17]

Answer: 0.024\ \text{miles per sec}

Step-by-step explanation:

Given

Path is changing according to the function f(x)=0.03(10x-x^2)

Rate of change of the elevation is given by the derivative of the function

\Rightarrow f'(x)=0.03(10-2x)

At x=1\ f'(x) \ \text{is}

\Rightarrow f'(x=1)=0.03(10-2\times 1)\\\Rightarrow f'(x=1)=0.03(8)\\\Rightarrow f'(x=1)=0.24\ \text{miles per sec}

3 0
2 years ago
If point d is placed on AC, how will the measure of DAB relate to the measure of CAB?
Lostsunrise [7]

angle DAB will be equal to angle CAB

4 0
2 years ago
Read 2 more answers
[Pre-Calc] Please Help! I don’t know where to start. How do I do this?
sertanlavr [38]

Answer:

See Below.

Step-by-step explanation:

Problem A)

We have:

\displaystyle \csc^2\theta \tan^2\theta -1=\tan^2\theta

When in doubt, convert all reciprocal trig functions and tangent into terms of sine and cosine.

So, let cscθ = 1/sinθ and tanθ = sinθ/cosθ. Hence:

\displaystyle \left(\frac{1}{\sin^2\theta}\right)\left(\frac{\sin^2\theta}{\cos^2\theta}\right)-1=\tan^2\theta

Cancel:

\displaystyle \frac{1}{\cos^2\theta}-1=\tan^2\theta

Let 1/cosθ = secθ:

\sec^2\theta -1=\tan^2\theta

From the Pythagorean Identity, we know that tan²θ + 1 = sec²θ. Hence, sec²θ - 1 = tan²θ:

\tan^2\theta =\tan^2\theta

Problem B)

We have:

\sin^3x=\sin x-\sin x \cos^2 x

Factor out a sine:

\sin x(\sin^2 x)=\sin x-\sin x\cos^2 x

From the Pythagorean Identity, sin²θ + cos²θ = 1. Hence, sin²θ = 1 - cos²θ:

\sin x(1-\cos^2 x)=\sin x-\sin x\cos^2x

Distribute:

\sin x- \sin x \cos^2 x=\sin x-\sin x\cos^2 x

Problem C)

We have:

\displaystyle \frac{\cos 2x+1}{\sin 2x}=\cot x

Recall that cos2θ = cos²θ - sin²θ and that sin2θ = 2sinθcosθ. Hence:

\displaystyle \frac{\cos^2 x-\sin^2 x+1}{2\sin x\cos x}=\cot x

From the Pythagorean Identity, sin²θ + cos²θ = 1 so cos²θ = 1 - sin²θ:

\displaystyle \frac{2\cos^2 x}{2\sin x\cos x}=\cot x

Cancel:

\displaystyle \frac{\cos x}{\sin x}=\cot x

By definition:

\cot x = \cot x

3 0
2 years ago
PLEASE HELP ME I WILL GIVE BRAINLIEST!!!!
Tju [1.3M]

Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation:

7 0
3 years ago
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