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brilliants [131]
3 years ago
7

Nola purchased two. 5 pounds of cheese for 10. $50 her mother purchased 3 pounds of the same cheese for $12.60 the cost of the c

heese varies directly with the number of pounds purchased how much does 1 pound of cheese cost
Mathematics
1 answer:
Dovator [93]3 years ago
7 0

Answer: 1 pound of cheese cost $4.2

Step-by-step explanation:

Nola purchased 2.5 pounds of cheese for 10. $50

Her mother purchased 3 pounds of the same cheese for $12.60

If the cost of the cheese varies directly with the number of pounds purchased, it means that the higher the pounds of cheese purchased, the higher the cost

To determine the the cost of 1 pound of cheese, we wound divide the the cheese by the number of pounds bought.

It becomes 10.5/2.5 = 4.2

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1. y = x2 + 8x + 15<br> Find the zeros of the function by rewriting the function in intercept form
lubasha [3.4K]

The zeros of given function y=x^{2}+8 x+15 is – 5 and – 3

<u>Solution:</u>

\text { Given, equation is } y=x^{2}+8 x+15

We have to find the zeros of the function by rewriting the function in intercept form.

By using intercept form, we can put value of y as  to obtain zeros of function

We know that, intercept form of above equation is x^{2}+8 x+15=0

\text { Splitting } 8 x \text { as }(5+3) x \text { and } 15 \text { as } 5 \times 3

\begin{array}{l}{\rightarrow x^{2}+(5+3) x+5 \times 3=0} \\\\ {\rightarrow x^{2}+5 x+3 x+5 \times 3=0}\end{array}

Taking “x” as common from first two terms and “3” as common from last two terms

x (x + 5) + 3(x + 5) = 0

(x + 5)(x + 3) = 0

Equating to 0 we get,

x + 5 = 0 or x + 3 = 0

x = - 5 or – 3

Hence, the zeroes of the given function are – 5 and – 3

5 0
3 years ago
Ok this should be easy:]
Blizzard [7]

Answer:

this is 8 raise to power 18

7 0
3 years ago
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2 gallons : 31 miles
Vika [28.1K]
15.5, because 31 divided by 2 is 15.5.  And all the other number divided by the number below them also turns out to be 15.5. 

Hope this helps. :)

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3 years ago
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3 years ago
Solve 2 sec^2 x + tan^2 x - 3=0.
Natasha2012 [34]
<span>2 sec^2 x + tan^2 x - 3 = 0
2(1 + tan^2 x) + tan^2 x - 3 = 0              [sec^2 x = 1 + tan^2 x]
2 + 2tan^2 x + tan^2 x - 3 = 0
3tan^2 x = 1
tan^2 x = 1/3

\tan x = \pm \sqrt{\frac{1}{3}} =\frac{1}{\sqrt{3}} \\ x=\tan^{-1}\frac{1}{\sqrt{3}}= \frac{\pi}{6} \\ =n \pi-\frac{5\pi}{6},\ \ n\pi-\frac{\pi}{6}</span><span />
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3 years ago
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