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Grace [21]
3 years ago
5

HELP ME PLEASE! 30 points

Mathematics
2 answers:
exis [7]3 years ago
7 0

Answer:

C=14

Step-by-step explanation:

To find the minimum value, graph each of the inequalities. After graphing each inequality, test a point and shade the region that satisfies the inequality. Once all inequalities have been shaded, find the region where they all overlap. The region will be bounded by intersection points. Test each of these points into C=x+3y. The least value for C is the minimum.

(14,0)                  (0,17.5)                   (3.08,3.64)

C=14+3(0)           C=0+3(17.5)           C=3.08 + 3(3.64)

C=14                   C=52.5                   C=14

kkurt [141]3 years ago
6 0

Hello there,

Answer: C=14



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tia_tia [17]

a 10-by-10 grid means there are 100 squares and percent is out of 100 so if you have 42% then you just need to fill in 42 squares

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3 years ago
8÷2 (2+2)=X <br> What is the value of x?
faltersainse [42]
8/2(2+2)=X
8/2(4)=X
4(4)=X
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6 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B5x%2F8y%7D" id="TexFormula1" title="\sqrt[4]{5x/8y}" alt="\sqrt[4]{5x/8y}" al
Furkat [3]

Answer:  \frac{\sqrt[4]{10xy^3}}{2y}

where y is positive.

The 2y in the denominator is not inside the fourth root

==================================================

Work Shown:

\sqrt[4]{\frac{5x}{8y}}\\\\\\\sqrt[4]{\frac{5x*2y^3}{8y*2y^3}}\ \ \text{.... multiply top and bottom by } 2y^3\\\\\\\sqrt[4]{\frac{10xy^3}{16y^4}}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{16y^4}} \ \ \text{ ... break up the fourth root}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{(2y)^4}} \ \ \text{ ... rewrite } 16y^4 \text{ as } (2y)^4\\\\\\\frac{\sqrt[4]{10xy^3}}{2y} \ \ \text{... where y is positive}\\\\\\

The idea is to get something of the form a^4 in the denominator. In this case, a = 2y

To be able to reach the 16y^4, your teacher gave the hint to multiply top and bottom by 2y^3

For more examples, search out "rationalizing the denominator".

Keep in mind that \sqrt[4]{(2y)^4} = 2y only works if y isn't negative.

If y could be negative, then we'd have to say \sqrt[4]{(2y)^4} = |2y|. The absolute value bars ensure the result is never negative.

Furthermore, to avoid dividing by zero, we can't have y = 0. So all of this works as long as y > 0.

3 0
3 years ago
Let m Z1=(3x)º, m 22=(5x-7)", m 23=(4x+15), and m ZAFD=128º. Find m44 and m ZBFD.​
olga55 [171]

Answer:

m<4 = 52°

m<BFD = 98°

Step-by-step explanation:

m<1 = (3x)°

m<2 = (5x - 7)°

m<3 = (4x + 15)°

m<AFD = 128°

✔️Find m<4:

m<4 = 180° - m<AFD (angles on a straight line)

Substitute

m<4 = 180° - 128°

m<4 = 52°

✔️m<BFD = m<2 + m<3

Substitute

m<BFD = (5x - 7)° + (4x + 15)°

We need to find the value of x.

Create an equation to find x.

m<1 + m<2 + m<3 = m<AFD (angle addition postulate)

Substitute

3x + 5x - 7 + 4x + 15 = 128°

Add like terms and solve for x

12x + 8 = 128

12x + 8 - 8 = 128 - 8

12x = 120

12x/12 = 120/12

x = 10

m<BFD = (5x - 7)° + (4x + 15)°

Plug in the value of x

m<BFD = 5(10) - 7 + 4(10) + 15

m<BFD = 50 - 7 + 40 + 15

m<BFD = 98°

4 0
3 years ago
In the expansion of (3a + 4b)8, which of the following are possible variable terms? Explain your reasoning. a2b3; a5b3; ab8; b8;
Fiesta28 [93]

We have to find the expansion of (3a+4b)^{8}

We will use binomial expansion to expand the given expression, which states that the expression (a+b)^{n} is expanded as :

(a+b)^{n}=^{n}C_{0}a^{n}+^{n}C_{1}a^{n-1}b+^{n}C_{2}a^{n-2}b^{2}+........^{n}C_{n}b^{n}

Now expanding (3a+4b)^{8} we get,

(3a+4b)^{8}=^{8}C_{0}(3a)^{8}+^{8}C_{1}(3a)^{7}(4b)+^{8}C_{2}(3a)^{6}(4b)^{2}+^{8}C_{3}(3a)^{5}(4b)^{3}+^{8}C_{4}(3a)^{4}(4b)^{4}+^{8}C_{5}(3a)^{3}(4b)^{5}+^{8}C_{6}(3a)^{2}(4b)^{6}+^{8}C_{7}(3a)(4b)^{7}+^{8}C_{8}(4b)^{8}

(3a+4b)^{8}=^{8}C_{0}(3)^{8}a^{8}+^{8}C_{1}(3)^{7}(4)(a^{7}b)+^{8}C_{2}(3)^{6}(4)^{2}(a^{6}b^{2})+^{8}C_{3}(3)^{5}(4)^{3}(a^{5}b^{3})+^{8}C_{4}(3)^{4}(4)^{4}(a^{4}b^{4})+^{8}C_{5}(3)^{3}(4)^{5}(a^{3}b^{5})+^{8}C_{6}(3)^{2}(4)^{6}(a^{2}b^{6})+^{8}C_{7}(3)(4)^{7}(ab^{7})+^{8}C_{8}(4)^{8}(b^{8})

So, the variables are a^{5}b^{3} , b^{8} , a^{4}b^{4} , a^{8}  , [tex] ab^{7}

5 0
3 years ago
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