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mylen [45]
3 years ago
5

House prices in a certain area went

Mathematics
1 answer:
Scorpion4ik [409]3 years ago
8 0

Answer:

£60000

Step-by-step explanation:

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Evaluate each expression for g=10 9-4g+1=?
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-30, 10*4=40 so you do 9-40+1 which is equal to -30
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Patsy found a top on the clearance rack marked 60% off the original price. If the original price is $48, how much will Patsy sav
Arlecino [84]
If the top is $48 and it is 60% off you will multiply the cost by the decimal form of the sale percentage: 48*0.60= $28.80. This is the amount she will save and I think is the answer you are looking for.

If there is a follow up question asking for the final cost of the top you would find that by subtracting the amount saved from the original cost: 48.00-28.80=$19.20.
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Find the cosine of each acute angle
Lemur [1.5K]

Answer:

cos P = 53 degrees

cos Q = 37 degrees

Step-by-step explanation:

cos P = \frac{adjacent}{hypothenuse}

cos P = \frac{24}{40}

cos P = 0.60

Cos-1 of 0.60 is 53.13 (round to the nearest whole number)

cos P = 53 degrees

cos Q = \frac{adjacent}{hypothenuse}

cos Q = \frac{32}{40}

cos Q = 0.80

Cos of 0.80 is 36.86 (round to the nearest whole number)

cos Q = 37 degrees

To check anser:

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4 0
3 years ago
The stem and lead plot shows data from a set of size 80 what are the median and mode
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Are there any more numbers to this problem ? a picture of the data
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A pumpkin is launched from the top of a 20 foot tall platform at an initial velocity of 84 feet per second. The height, h, of th
Gennadij [26K]
<h2>Explanation:</h2>

In this problem we have a pumpkin is launched from the top of a 20 foot tall platform at an initial velocity of 84 feet per second. So the height, h, of the pumpkin at time t seconds after the launch can be modeled by the equation:

h(t) = -16t^2 + 84t + 20

So this is the equation of a parabola. The maximum of this parabola occurs at its vertex. So let's find this vertex:

\text{The x-value of the vertex is}: \\ \\ t=-\frac{b}{2a} \\ \\ \\ Where: \\ \\ a=-16 \\ \\ b=84 \\ \\ c=20 \\ \\ \\ t=-\frac{84}{2(-16)} \\ \\ t=-\frac{84}{-32} \\ \\ t=\frac{21}{8}=2.625 \\ \\ h(2.625)=-16(2.625)^2+84(2.625)+20 \\ \\ h(2.625)=-110.25+220.5+20 \\ \\ h(2.625)=130.25m

Finally, the maximum occurs at time 2.625 seconds when the height is 130.25m

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