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lisabon 2012 [21]
4 years ago
13

Max bought a package of hamburger that weighs 5 1/3 pounds. He needs 1/3 of that amount for a recipe.

Mathematics
1 answer:
boyakko [2]4 years ago
5 0
1.5 pounds would be the answer,
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Could 10.3cm 4.4cm and 8.3 cm be the side lengths of a triangle yes or no
evablogger [386]

Answer:

No

Step-by-step explanation:

As you probably know, triangles have 3 sides. The longest side is called the hypotenuse. In this case, the hypotenuse is 10.3. Now, you might or might not know the pythagorean theorem, which states that the <em>a² + b² =c ². </em>

In this case, we can say that 4.4 is <em>a </em>and 8.3 is <em>b. </em>Now if we square 4.4 and add it to 8.3 squared, we get 88.25. However, if you square 10.3, you get 106.09. Thus, the values cannot be this way. So, your answer is no.

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3 years ago
Cube A and Cube B are similar solids. the volume of cube A is 27 cubic inches , and the volume of cube B is 125 cubic inches. ho
Norma-Jean [14]
The answer should be C because that’s the most reasonable and because you can’t simplify 125/7
3 0
4 years ago
Read 2 more answers
Help appreciated! (Will mark as Brainliest)
WARRIOR [948]

Answer:

trapezoid

Step-by-step explanation:

parallelograms require two sets of parallel lines while trapezoids do not, only requiring one set. it is possible to get a trapezoid with 2 90 degree angles.

3 0
3 years ago
Sharon made a scatterplot comparing the shoulder heights of dogs to their weights.
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Answer:

105

Step-by-step explanation:

6 0
3 years ago
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a plane flying with the wind travels 630 mi in 1.5hrs. Flying against the wind , the plane travels 1120 mi in 4hrs. Find the rat
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\bf \begin{cases}&#10;r=\textit{rate of the plane}\\&#10;w=\textit{rate of the wind}\\&#10;\end{cases}&#10;\\\\&#10;&#10;\begin{array}{ccccllll}&#10;&distance&rate&time(hrs)\\&#10;&\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\\&#10;\textit{with wind}&630&r+w&1\frac{1}{2}\\&#10;\textit{against wind}&1120&r-w&4&#10;\end{array}&#10;

\bf thus  \begin{cases}&#10;630=(r+w)1\frac{1}{2}\to 630=(r+w)\frac{3}{2}&#10;\\\\&#10;1120=(r-w)4\\&#10;--------------\\&#10;630=(r+w)\frac{3}{2}\implies 630\cdot \frac{2}{3}=r+w\\&#10;420=r+w\implies \boxed{420-w}=r\\&#10;--------------\\&#10;thus\\&#10;--------------\\&#10;1120=(r-w)4\implies 1120=4r-4w&#10;\\\\&#10;1120=4(\boxed{420-w})-4w&#10;\end{cases}&#10;

solve for "w", to find the wind's speed rate,

so hmmm what's the plane's rate?  well 420 - w = r  :)
5 0
3 years ago
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