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3 years ago

How do I solve this?

Mathematics

1 answer:

Anni [7]3 years ago

3 0

I hope this helps you

4^3/3.x^2/3.y^3/3

4.y.x^2/3

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13x + 6y = -30

Brilliant_brown [7]

The answer is d, (-2.5,0.75)

5 0

3 years ago

Some body can help me with a geometric mean maze

Mars2501 [29]

Answer:

See explanation

Step-by-step explanation:

Theorem 1: The length of the altitude to the hypotenuse of a right triangle is the geometric mean of the lengths of the segments of the hypotenuse.

Theorem 2: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

1. Start point: By the 1st theorem,

$x^2=25\cdot (49-25)=25\cdot 24=5^2\cdot 2^2\cdot 6\Rightarrow x=5\cdot 2\cdot \sqrt{6}=10\sqrt{6}.$

2. South-East point from the Start: By the 2nd theorem,

$x^2=40\cdot (40+5)=4\cdot 5\cdot 2\cdot 9\cdot 5\Rightarrow x=2\cdot 5\cdot 3\cdot \sqrt{2}=30\sqrt{2}.$

3. West point from the previous: By the 2nd theorem,

$x^2=(32-20)\cdot 32=4\cdot 3\cdot 16\cdot 2\Rightarrow x=2\cdot 4\cdot \sqrt{6}=8\sqrt{6}.$

4. West point from the previous: By the 1st theorem,

$9^2=x\cdot 15\Rightarrow x=\dfrac{81}{15}=\dfrac{27}{5}=5.4.$

5. West point from the previous: By the 2nd theorem,

$10^2=8\cdot (8+x)\Rightarrow 8+x=12.5,\ x=4.5.$

6. North point from the previous: By the 1st theorem,

$x^2=48\cdot 6=6\cdot 4\cdot 2\cdot 6\Rightarrow x=6\cdot 2\cdot \sqrt{2}=12\sqrt{2}.$

7. East point from the previous: By the 2nd theorem,

$x^2=22.5\cdot 30=225\cdot 3\Rightarrow x=15\sqrt{3}.$

8. North point from the previous: By the 1st theorem,

$x^2=7.5\cdot 36=270\Rightarrow x=3\sqrt{30}.$

8. West point from the previous: By the 2nd theorem,

$x^2=12.5\cdot (12.5+13.5)=12.5\cdot 26=25\cdot 13\Rightarrow x=5\sqrt{13}.$

9. North point from the previous: By the 1st theorem,

$12^2=x\cdot 30\Rightarrow x=\dfrac{144}{30}=4.8.$

101. East point from the previous: By the 1st theorem,

$6^2=1.6\cdot (x-1.6)\Rightarrow x-1.6=22.5,\ x=24.1.$

11. East point from the previous: By the 2nd theorem,

$20^2=32\cdot (32-x)\Rightarrow 32-x=12.5,\ x=19.5.$

12. South-east point from the previous: By the 2nd theorem,

$18^2=x\cdot 21.6\Rightarrow x=15.$

13. North point=The end.

6 0

3 years ago

Complete the Domain and Range worksheet attached. You can type the answer into submit

VMariaS [17]

There different from there own way there celebrities and there ways

3 0

3 years ago

Which is equivalent to 3 cube root 125

aleksklad [387]

Do you mean $3 \sqrt[3]{125}$ or $\sqrt[3]{125}$
remember that $\sqrt[n]{x^m}=x^\frac{m}{n}$
resolve the $\sqrt[3]{125}$ part first $\sqrt[3]{125}=$ $\sqrt[3]{5^3}=$ $5^\frac{3}{3}=$ $5^1=$ $5$

so
$3 \sqrt[3]{125}=3*5=15$ or $\sqrt[3]{125} =5$ not sure which one you mean

7 0

3 years ago

Read 2 more answers

Graph: y< 1/3x<br> 2<br> Show Me<br> Draw

charle [14.2K]

Answer:

i dont understand please provide more info

Step-by-step explanation:

okay?

7 0

3 years ago