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maria [59]
3 years ago
9

A local grocery store is advertising a discount of 20% off a large bag of peaches, which normally costs $5.

Mathematics
2 answers:
mr_godi [17]3 years ago
8 0
20% of $5.00 is $1.00 Therefore your sale price is 1.00
Maksim231197 [3]3 years ago
4 0
I may be wrong, but 20% of $5 is $1, right? So, it's $1. 
Here's the work.

is/of = %/100

Take what you know and plug into the proportion. 
is/5 = 20/100
Cross multiply 5 and 20 and divide by 100
is = $1
20% of $5 = $1
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Please help me with this homework
soldi70 [24.7K]

Step-by-step explanation:

we see f(x), which is |x|.

and we see g(x), which is clearly the same basic graph, it is just shifted 3 units down in y direction.

shifts are simply done by keeping the original function definition and then add it subtract a certain constant that then adapts every original functional value.

so, a shift down by 3 units is done by adding -3.

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8 0
2 years ago
Y= 2.5x+ 5.8 when x = 0.6 pls find the function
adoni [48]
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3 years ago
When the value of t is 60, the value of c is 90 . Explain what this means using the problem context.
andrey2020 [161]

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3 years ago
What is the distance between the two end points (5,6) (-4,-7) round to the nearth two decimal places
garri49 [273]

The formula of a distance between two points:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

We have:

(5,\ 6)\to x_1=5,\ y_1=6\\(-4,\ -7)\to x_2=-4,\ y_2=-7

substitute:

d=\sqrt{(-4-5)^2+(-7-6)^2}=\sqrt{(-9)^2+(-13)^2}=\sqrt{81+169}=\sqrt{250}\\\\=\sqrt{25\cdot10}=\sqrt{25}\cdot\sqrt{10}=5\sqrt{10}

5 0
3 years ago
What is the exact value of cos (67.5°)?
babymother [125]

first off, make sure you have a Unit Circle, if you don't do get one, you'll need it, you can find many online.

let's double up 67.5°, that way we can use the half-angle identity for the cosine of it, so hmmm twice 67.5 is simply 135°, keeping in mind that 135° is really 90° + 45°, and that whilst 135° is on the 2nd Quadrant and its cosine is negative 67.5° is on the 1st Quadrant where cosine is positive, so

cos(\alpha + \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta) \\\\\\ cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}} \\\\[-0.35em] ~\dotfill\\\\ cos(135^o)\implies cos(90^o+45^o)\implies cos(90^o)cos(45^o)~~ - ~~sin(90^o)sin(45^o) \\\\\\ \left( 0 \right)\left( \cfrac{\sqrt{2}}{2} \right)~~ - ~~\left( 1\right)\left( \cfrac{\sqrt{2}}{2} \right)\implies -\cfrac{\sqrt{2}}{2} \\\\[-0.35em] ~\dotfill

cos(67.5^o)\implies cos\left( \frac{135^o}{2} \right)\implies \pm \sqrt{\cfrac{ ~~ 1-\frac{\sqrt{2} ~~ }{2}}{2}}\implies \stackrel{I~Quadrant}{+\sqrt{\cfrac{ ~~ 1-\frac{\sqrt{2} ~~ }{2}}{2}}} \\\\\\ \sqrt{\cfrac{ ~~ \frac{2-\sqrt{2}}{2} ~~ }{2}}\implies \sqrt{\cfrac{2-\sqrt{2}}{4}}\implies \cfrac{\sqrt{2-\sqrt{2}}}{\sqrt{4}}\implies \cfrac{\sqrt{2-\sqrt{2}}}{2}

8 0
2 years ago
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