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PSYCHO15rus [73]
2 years ago
8

Solve 12xx + 16 - 1 = 0​

Mathematics
1 answer:
zhuklara [117]2 years ago
5 0

Answer:

xx=-0.8

Step-by-step explanation:

not sure if this is right because the 12xx is kind of strange

12xx+16-1=0
12xx+15=0
12xx=-15
12/12xx=12/-15
xx=-.8

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9/11 = ?/22<br> a. 12<br> b. 9<br> c. 4<br> d. 18
svlad2 [7]
18/22 = 2*9/2*11 = 9/11

Therefore, your answer is 18
7 0
3 years ago
Please help 50 points + Brainliest
Talja [164]

Step-by-step explanation:

AB · BE = CB · BD            Given

CB/BE = AB/BD               Division Property of Equality

<ABC = <DBE                  Vertical angles are equal

ΔABC ≅ ΔDBE                SAS Similarity Theorem


Hope it helps

7 0
3 years ago
Read 2 more answers
What is the area of triangle pqr on the grid? a triangle pqr is shown on a grid. the vertex p is on ordered pair 7 and 6, vertex
horsena [70]

P(7,6), Q(1,6), R(4,2)

We have PQ parallel to the x axis. We'll call that the base,

b = 7 - 1 = 6

The altitude is then the y difference h = 6 - 2 = 4

The area is \frac 1 2 bh = \frac 1 2 (6)(4) = 12

Answer: 12 square units


In general we can use the shoelace formula for the area of any polygon given coordinates. We write the points like this:

(7,6), (1,6), (4,2)

(1,6), (4,2), (7,6)

The area is then half the absolute value of the sum of the cross products:

A = \frac 1 2 | 7(6)-6(1) + 1(2)-6(4) + 4(6)-2(7) | = \frac 1 2 |24| = 12


6 0
3 years ago
Read 2 more answers
If m FDG = 50°, what is m FEG?<br><br> 75°<br> 50°<br> 100°<br> 25°<br><br> Thank you!
denpristay [2]
The measure of the angle FEG is probably the same as the measure of the angle FDG. There is nothing saying that EF is parallel to DG, but if so, the measure of the angle FEG is also 50º.
When the diagonals of a trapezium with two parallel bases inside of a circle are drawn, they make the same angle measure with those bases.
7 0
3 years ago
Please help, I have been trying for a while :(<br><br>question on photo​
Zigmanuir [339]

Answer:

\frac{x-22}{(x+2)(x-4)}

Step-by-step explanation:

multiply the numerator/denominator of the first fraction by (x - 4)

multiply the numerator/denominator of the second fraction by (x + 2)

this ensures that the fractions have a common denominator

\frac{4}{x+2} - \frac{3}{x-4}

= \frac{4(x-4)}{(x+2)(x-4)} - \frac{3(x+2)}{(x+2)(x-4)} ← subtract numerators leaving the common denominator

= \frac{4x-16-3x-6}{(x+2)(x-4)}

=\frac{x-22}{(x+2)(x-4)}

4 0
2 years ago
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