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ra1l [238]
3 years ago
7

Which expression has a value of 9/10

Mathematics
1 answer:
Luda [366]3 years ago
7 0

Answer:

Me too

Step-by-step explanation:

Is there a good

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If the slope of a line is -1/4,<br>find x when (x,2) and (1/2,6)​
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Use slope formula:
m = (y2-y1) / (x2-x1)
-1/4 = 6-2 / 1/2 - x
-1/4 = 4 / 1/2 - x
-1(1/2 - x) = 4(4)
-1/2 + x = 16

x = 16 1/2 or 33/2
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What number is smaller number -5 or -15? Explain your answer. Please help college math basic algebra
Goryan [66]
-15 is the smaller number because the further you go into the negatives like -30, -47, -52; youre getting further away from 0 and after 0 are positive numbers so -15 would be the smaller number
8 0
3 years ago
Read 2 more answers
Help plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
Setler79 [48]

Answer:

diameter = 3,600 units

Step-by-step explanation:

Let \pi = 3.14

C = 2\pir

11,304 = 2(3.14)r

r = 1,800

d = 2r = 2(1,800) = 3,600

7 0
2 years ago
How many different combinations are possible if each lock contains the numbers 0 to 39, and each combination contains three dist
Georgia [21]
(e) Each license has the formABcxyz;whereC6=A; Bandx; y; zare pair-wise distinct. There are 26-2=24 possibilities forcand 10;9 and 8 possibilitiesfor each digitx; yandz;respectively, so that there are 241098 dierentlicense plates satisfying the condition of the question.3:A combination lock requires three selections of numbers, each from 1 through39:Suppose that lock is constructed in such a way that no number can be usedtwice in a row, but the same number may occur both rst and third. How manydierent combinations are possible?Solution.We can choose a combination of the formabcwherea; b; carepair-wise distinct and we get 393837 = 54834 combinations or we can choosea combination of typeabawherea6=b:There are 3938 = 1482 combinations.As two types give two disjoint sets of combinations, by addition principle, thenumber of combinations is 54834 + 1482 = 56316:4:(a) How many integers from 1 to 100;000 contain the digit 6 exactly once?(b) How many integers from 1 to 100;000 contain the digit 6 at least once?(a) How many integers from 1 to 100;000 contain two or more occurrencesof the digit 6?Solutions.(a) We identify the integers from 1 through to 100;000 by astring of length 5:(100,000 is the only string of length 6 but it does not contain6:) Also not that the rst digit could be zero but all of the digit cannot be zeroat the same time. As 6 appear exactly once, one of the following cases hold:a= 6 andb; c; d; e6= 6 and so there are 194possibilities.b= 6 anda; c; d; e6= 6;there are 194possibilities. And so on.There are 5 such possibilities and hence there are 594= 32805 such integers.(b) LetU=f1;2;;100;000g:LetAUbe the integers that DO NOTcontain 6:Every number inShas the formabcdeor 100000;where each digitcan take any value in the setf0;1;2;3;4;5;7;8;9gbut all of the digits cannot bezero since 00000 is not allowed. SojAj= 9<span>5</span>
8 0
3 years ago
Someone please help or i’ll be failing geometry this year
Vlad [161]

Step-by-step explanation:

I don't know what constructions you were taught.

a "similar" triangle is a triangle with exactly the same angles as the other triangle, but the lengths of all sides are stretched or shortened by the same scaling factor f.

by saying 1:2 she means the second triangle should have sides with twice the lengths of the first triangle (f=2).

and the extra challenge - same basic thing. she allows you to pick one of the two triangles as reference. and then you need to draw a third triangle (again with the same angles) with the side lengths extended by the scaling factor f of 4/3.

I would draw the triangles right on top of each other with the same starting corner (let's call it A) for all 3.

we would get the triangles ABC, AMN and AXY.

the points B and C would be then halfway on AM and AN.

and M and N would then a bit before X and Y on AX and AY.

the beauty is, you only need to construct 2 sides of every new triangle down to the new endpoints. the third side is automatically scaled correctly, and you only need to connect these new endpoints.

let's assume you draw a triangle (just very simple) ABC with all side lengths being 3. so, AB=3, AC=3, BC=3.

now you draw AMN by extending AB and AC to a side length of 6 (f=2) creating M and N, and you connect M and N.

and then you can create the third triangle AXY by extending AM and AN by a factor of 4/3 to side lengths of 8 (4/3 × 6 = 8) creating new end points X and Y. and you connect X and Y.

and that is it. all 3 triangles are similar (the same angles), and all sides of a triangle have the same length ratio to the sides of the other triangle(s).

4 0
3 years ago
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