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Reil [10]
4 years ago
13

Assinale todas as assertivas corretas com relação aos princípios de ensino na Matemática.

Mathematics
1 answer:
Neko [114]4 years ago
6 0
Do u speak English...............
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Rebecca has 1 2 of a yard of fabric. she divides the fabric into 9 equal sized strips for an art project. how many yards of fabr
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It will be 108 i think.

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3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!
elena-s [515]

<u>"An on-line electronics store must sell </u><u>at least $5000</u><u> worth of computers and printers per day"</u>

The bold faced words tell us greater than or equal to 5000


<u>"The store can ship a </u><u>maximum of 20</u><u> items per day"</u>

The bold faced words tell us less than or equal to 20


We can automatically eliminate A and B since they don't have EQUAL sign in them. From C and D, we choose D because it correctly attaches the price of printers and computers to the respective variables <em>(printers is p & computers is q)</em>. Choice C does this wrong!


ANSWER: D

6 0
3 years ago
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Find the measure of Angle F ASAP thank u !!
KATRIN_1 [288]

Answer:

its 40

Step-by-step explanation: 65+75= 140     180-140=40

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5 0
3 years ago
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In T-ball, the distance to each successive base is 50 feet. If the distance from home plate to the pitcher’s mound is 38 feet, h
JulijaS [17]

Answer:

<h2>The distance from the pitcher's mound and to second base is 37.99 approximately.</h2>

Step-by-step explanation:

The diamond is a square, which in this case has 50 feet long each side, and from home to pitcher is 38 feet. Notice that home is a vertex of the square and the pitcher's mound is the intersection of the diagonals, where they cut half.

We can find the distance from the pitcher to first base using Pythagorean's Theorem, where 50 feet is the hypothenuse.

50^{2} =38^{2}+x^{2}\\x^{2}=50^{2}-38^{2}\\x=\sqrt{2500-1444}\\ x=\sqrt{1056}\\ x \approx 32.5 \ ft

Therefore, the distance from the pitcher to first base is 32.5 feet, approximately.

Now, we can use again Pythagorean's Theorem to find the distance from pitcher to second base, where the hypothenuse is 50 feet.

50^{2}=32.5^{2}+y^{2}\\y^{2}=50^{2}-32.5^{2}\\y=\sqrt{2500-1056.25}\\ y =\sqrt{1443} \approx 37.99

Therefore, the distance from the pitcher's mound and to second base is 37.99 approximately.

<em>(this results make sense, because the diagonals of a square intersect at half, that means all bases have the same distance from pitcher's mound, so the second way to find the distance asked in the question is just using theory)</em>

8 0
3 years ago
Please helppppp the thing has to be 20 characters long soooo how was your day today?? Good, bad? I hope it was good!
prohojiy [21]

Answer: (4, -1) and (4, -5)

Step-by-step explanation:

The x coordinate is 4 and the y coordinate is -1 and -5. The endpoints are those two closed/bulleted dots on the graph.

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3 years ago
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