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sweet [91]
3 years ago
8

What's the answer to the picA. 3 2/3B. 2 2/3C. 3 1/3D. -7 1/3 ​

Mathematics
2 answers:
ElenaW [278]3 years ago
8 0
The answer to the question is: 2 2/3
ICE Princess25 [194]3 years ago
4 0

Answer: 2 2/3

Step-by-step explanation:

-2 1/3-(-5)

-7/3-(-5)

8/3

2 2/3

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you can buy a 48​-pound bag of flour for ​$11 or you can buy a​ 1-pound bag for  ​$0.49. Compare the per pound cost for the larg
Pavel [41]
$48÷$11=$4.36 repeated

while buying it the other way will save you way more money with $0.49 per pound

5 0
3 years ago
Read 2 more answers
In triangles GHI and RST, ∠G ≅ ∠R, ∠H ≅ ∠S, and segment GI ≅ segment RT. Is this information sufficient to prove triangles GHI a
user100 [1]

The given information is not sufficient to prove ΔGHI and ΔRST are congruent through SSS congruence.

<h3>What is SSS congruence of triangles?</h3>

As per the SSS congruence of a triangle, if all the three sides of a triangle are equal to the all the three corresponding sides of another triangle, then the two triangles are congruent triangle.

Since one side and two angles of the triangle are given, the triangles can not be proved to be congruent as per the SSS congruence of triangles. Hence, the given information is not sufficient to prove ΔGHI and ΔRST are congruent through SSS congruence.

Learn more about SSS Congruence:

brainly.com/question/535562

#SPJ1

3 0
2 years ago
A system for tracking ships indicated that a ship lies on a hyperbolic path described by 5x2 - y2 = 20. the process is repeated
zysi [14]
Answer:
The ship is located at (3,5)

Explanation:
In the first test, the equation of the position was:
5x² - y² = 20 ...........> equation I
In the second test, the equation of the position was:
y² - 2x² = 7 ..............> equation II
This equation can be rewritten as:
y² = 2x² + 7 ............> equation III

Since the ship did not move in the duration between the two tests, therefore, the position of the ship is the same in the two tests which means that:
equation I = equation II

To get the position of the ship, we will simply need to solve equation I and equation II simultaneously and get their solution.

Substitute with equation III in equation I to solve for x as follows:
5x²-y² = 20
5x² - (2x²+7) = 20
5x² - 2y² - 7 = 20
3x² = 27
x² = 9
x = <span>± </span>√9

We are given that the ship lies in the first quadrant. This means that both its x and y coordinates are positive. This means that:
x = √9 = 3

Substitute with x in equation III to get y as follows:
y² = 2x² + 7
y² = 2(3)² + 7
y = 18 + 7
y = 25
y = +√25
y = 5

Based on the above, the position of the ship is (3,5).

Hope this helps :)
8 0
3 years ago
Olympic volleyball player Bev Oden averaged 3.42 "kills" (spikes that end a rally) per game. Predict her number of kills in a 5
damaskus [11]

Answer: 17.1

Step-by-step explanation:

If she averages 3.42 kills per game, she should average that for each of the 5 games. You need to multiply 3.42 times 5 for the total of 5 games.

3.42 x 5 = 17,1

7 0
4 years ago
Show that ( 2xy4 + 1/ (x + y2) ) dx + ( 4x2 y3 + 2y/ (x + y2) ) dy = 0 is exact, and find the solution. Find c if y(1) = 2.
fredd [130]

\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}

\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}

so the ODE is indeed exact and there is a solution of the form F(x,y)=C. We have

\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)

\dfrac{\partial F}{\partial y}=4x^2y^3+\dfrac{2y}{x+y^2}=4x^2y^3+\dfrac{2y}{x+y^2}+f'(y)

f'(y)=0\implies f(y)=C

\implies F(x,y)=x^2y^3+\ln(x+y^2)=C

With y(1)=2, we have

8+\ln9=C

so

\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}

8 0
3 years ago
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