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Rina8888 [55]
3 years ago
14

Find the y intercept of the graph

Mathematics
1 answer:
Dmitriy789 [7]3 years ago
5 0

Answer:

i believe the y intersept is -4

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a rectangular picture frame has a width of 9 cm and a length of x cm. which expression represents the perimeter of the picture f
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The phone company Blurizon has a monthly cellular plan where a customer pays a flat fee for unlimited voice calls and then a cer
emmainna [20.7K]

Answer:

y=2x+27.

Step-by-step explanation:

We have been given that when a customer uses 14 GB, the monthly cost will be $55. When the customer uses 38 GB, the monthly cost will be $103. We are asked to write an equation in slope-intercept form, where x is the number of GB of data used in a month and y is the total monthly cost of the Blurizon plan.

We have been given two points on line (14,55) and (38,103).  

Let us find slope of line.

m=\frac{103-55}{38-14}

m=\frac{48}{24}

m=2

Now, we will use slope-intercept form of equation.

y=mx+b

55=2(14)+b

55=28+b

55-28=28-28+b

b=27

Therefore, our required equation would be y=2x+27.

6 0
4 years ago
A cone has a radius of 5 cm and a height of 9 cm. What is the volume of the cone to the nearest whole number? Use 3.14 for pi.
Katyanochek1 [597]

Answer:

235or 236

Step-by-step explanation:

A cone has a radius of 5 cm

a height of 9 cm

Volume of cone =1/3πr^2h

=1/3×3.14×5×5×9

= 3.14×25×3

=3.14×75

=235.5

=235or 236 is the answer

8 0
3 years ago
How many half-lives of radon-222 have passed in 11.46 days? If 5.2 × 10−8 g of radon-222 remain in a sealed box after 11.46 days
Svetach [21]

Answer:

(i) Approximately 3 half lifes

(ii) 4.21\times 10^{-7}\text{ g}

Step-by-step explanation:

(i) ∵ The half life of Radon-222 is approximately 3.8 days,

So, the number of half life in 11.46 days = \frac{11.46}{3.8} ≈ 3

(ii) Since, the half life formula is,

N=N_0 (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}

Where,

N_0 = initial quantity,

t = number of periods

t_{\frac{1}{2}} = half life of the quantity,

Given,

N = 5.2\times 10^{-8}\text{ g}

t = 11.46 days,

t_{\frac{1}{2}} = 3.8\text{ days}

\implies 5.2\times 10^{-8}=N_0 (\frac{1}{2})^\frac{11.46}{3.8}

\implies N_0=2^{\frac{11.46}{3.8}}\times 5.2\times 10^{-8}\approx 4.21\times 10^{-7}\text{ g}

7 0
4 years ago
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