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STatiana [176]
3 years ago
9

What is the solution set for the open sentence, using the given replacement set ?

Mathematics
1 answer:
77julia77 [94]3 years ago
4 0

Answer:

x = 3

Step-by-step explanation:

first move 7 into RHS in addition

then move t into RHS in division

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Use trigonometry please
Elena-2011 [213]
Answer is 

tan º = 16/x
16/tan 33=x 
x=24.637

8 0
4 years ago
Triangle XYZ has been enlarged and P as the center of dilation and scale to form a triangle X’ Y’ Z’
Maksim231197 [3]

Distances from the center of dilation are multiplied by the dilation factor. The appropriate choice is ...

... A. X'P/XP = Y'P/YP = Z'P/ZP

_____

We don't know what selection B means in this context.

Selection C will only be true if the dilation is by a factor of 2.

Selection D will only be true if P is the center of the circumscribing circle for both X'Y'Z' and XYZ.

6 0
4 years ago
Divide the given polynomial by the given monomial 3 x square - 2 x divided by X ​
inna [77]
Dividing (3x^2-2x) by x gives out 3x-2
7 0
3 years ago
The blueprint of a concrete patio has a scale of 2 in. = 3 ft You want to find the dimensions of a new blueprint of the patio wi
LenKa [72]

Answer:

If\ old\ scale\ is\ taken:\ Length=45\ ft,\ Width=42\ ft\\\\If\ new\ scale\ is\ taken:\ Length=36\ ft,\ Width=33.6\ ft

Step-by-step explanation:

Older Scale:

2\ in=3\ ft\\\\\Rightarrow 1\ in=\frac{3}{2}\ ft\\\\Length=30\ in\\\\Length=30\times \frac{3}{2}\ ft\\\\Length=45\ ft\\\\Width=28\ in\\\\width=28\times \frac{3}{2}\ ft\\\\width=42\ ft

New scale:

4\ in=5\ ft\\\\\Rightarrow 1\ in=\frac{5}{4}\ ft\\\\Length=30\ in\\\\Length=30\times \frac{5}{4}\ ft\\\\Length=36\ ft\\\\Width=28\ in\\\\width=28\times \frac{5}{4}\ ft\\\\width=33.6\ ft

4 0
3 years ago
there are 5 more than twice as many students taking Algebra 1 then taking algebra 2. If there are 44 students taking Algebra 2,
Temka [501]

Answer:

93

Step-by-step explanation:

Key :

A1 = Algebra 1

A2 = Algebra 2

Alright so basically lets first look at the info they gave us :

We have 5 more than twice as many students taking A1 than we do A2.

We have 44 students taking A2.

And we need to find the least amount of students that could be taking A1.

So we need to take the amount of students taking A2 (44) and double it to find the amount taking A1.

So we can do 44 x 2 = 88 to get this.

But the problem also states there is 5 more then twice the number of students taking A2.

So we have that 88 but now we just need to add 5 to make up for them telling us that in the problem.

So :

88 + 5 = 93

Our final answer and least amount of students taking A1 is 93 students.

8 0
3 years ago
Read 2 more answers
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