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Tomtit [17]
3 years ago
9

1/2x + 5y- 10< 0 The point (4,y) is a solution for the inequality shown. What is the value for y.

Mathematics
2 answers:
finlep [7]3 years ago
8 0
Subtract x on both sides the result you divide it by 5
Phoenix [80]3 years ago
7 0
1/2×4+5y-10 <0
2+5y-10 <0
5y-8 <0
+8 +8
5y <8
y <1.6
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A pair of jeans originally cost $80, but is on sale for 30% off. there is a coupon available for an additional 10% off of the sa
Lady_Fox [76]
30% = 0.3. 

0.3*80 = 24, so we know that 30% of 80 is 24. 

80 - 24 = 56, so after the sale, the price for a pair of jeans is $56.

10% = 0.1

0.1*56 = 5.6, so with the coupon, we get $5.60 off. 

56 - 5.60 = 50.40

The jeans cost $50.40 after the sale and the coupon. 
5 0
3 years ago
27^{3x}=(1/9)^x<br> write in b^x=b^y form
iogann1982 [59]

Answer:

3^9x=3^-2x

Step-by-step explanation:

27^3x=(1/9)^x

3^3×3x=(1/3^2)^x

3^9x=3^-2x

5 0
3 years ago
At a hot dog stand they serve regular hot dogs and footling hot dogs at a ratio of 5 to 3. Based on this ratio how many footlong
natta225 [31]
Answer:
45
Step-by-step explanation:
Ratio is 5:3
So total ratio "parts" is 5 + 3 = 8
In total there are 120 dogs served, so each part is:
120/8 = 15 dogs
Since, footlong hot dogs are "3" parts, and each part is 15 dogs, there will be:

3 * 15 = 45 footlong hotdogs
3 0
3 years ago
Can someone please help
mario62 [17]
The answer is B) Translation, Congruent
6 0
3 years ago
The equation to model Exponential Growth is:
abruzzese [7]

Your salary in x years is modeled an the exponential growth

The equation that determines your salary in x years is y = 45000(1.05)^x

<h3>How to model the salary growth?</h3>

The model of the exponential growth is given as:

y = a(1 + r)^x

From the question, we have:

Initial salary, a = 45000

Raise, r = 5%

So, the equation becomes

y = 45000(1 + 5%)^x

Evaluate the sum

y = 45000(1.05)^x

Hence, the equation that determines your salary in x years is y = 45000(1.05)^x

Read more about exponential functions at:

brainly.com/question/11464095

5 0
2 years ago
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