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Sergeu [11.5K]
3 years ago
10

Find the term that must be added to the equation x2−10x=1 to make it into a perfect square

Mathematics
1 answer:
adoni [48]3 years ago
6 0

Answer:

25

Step-by-step explanation:

If you add 25 to both sides, you get

x^2-10x+25=26.

Now, you can simplify the left side into a perfect square

(x-5)^2=26

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Check all the ordered pairs that satisfy the equation below. Y=6x
slava [35]
If \frac{y}{x} = 6, then any ordered pair would satisfy the equation.
Since we're not given the ordered pairs, then this is the only information I can give to you.
3 0
3 years ago
The high temperature for one town last year was 79°F The low temperature for the same town was - 3°F What was the difference bet
Ainat [17]

Answer:

79°c - (-3°c) = 82°c

Step-by-step explanation:

this is because we need to calculate the total gap between those numbers through number line. I have attached the photo, have a look over there.

6 0
3 years ago
HELp mE PLEASE FR NO CAP
Inessa [10]
First one is C,

Second is A

Third is C pretty sure

Fourth is D
4 0
3 years ago
Please help?!
Anton [14]

Answer:

Δ PQT ~ Δ QRS  .....{S-S-S test for similarity}...Proof is below.

Step-by-step explanation:

Given:

In Δ PQT

PQ = 30 ft

QT = 28 ft

TP = 20 ft

In Δ QRS

QR = 15 ft

RS = 14 ft

SQ = 10 ft

To Prove:

Δ PQT ~ Δ QRS

Proof:

First we consider  the ratio of the sides

\frac{PQ}{QR}=\frac{30}{15} = \frac{2}{1}            ..............( 1 )

\frac{QT}{RS}=\frac{28}{14} = \frac{2}{1}            ..............( 2 )

\frac{TP}{SQ}=\frac{20}{10} = \frac{2}{1}            ..............( 3 )

So By equation ( 1 ), ( 2 ) and  ( 3 ) we get

\frac{PQ}{QR}=\frac{QT}{RS} = \frac{TP}{SQ}

Now in Δ PQT  and Δ QRS we have

\frac{PQ}{QR}=\frac{QT}{RS} = \frac{TP}{SQ}

Which are corresponding sides of a similar triangle in proportion.

∴ Δ PQT ~ Δ QRS  .....{S-S-S test for similarity}...Proved

8 0
3 years ago
My brain stopped. Please help
Over [174]

Answer:

∠1 = 90°

∠2 = 66°

∠3 = 24°

∠4 = 24°

Step-by-step explanation:

Usually the diagonals of a rhombus bisect each other at right angles.

Thus; ∠1 = 90°

Since they bisect at right angles, then;

∠R1S = 90°

Now, sum of angles in a triangle is 180°

Thus;

66° + 90° + ∠4 = 180°

156 + ∠4 = 180

∠4 = 180 - 156

∠4 = 24°

Now, also in rhombus, diagonals bisect opposite angles.

Thus; ∠4 = ∠3

Thus, ∠3 = 24°

Similarly, the diagonal from R to T bisects both angles into 2 equal parts.

Thus; ∠2 = 66°

6 0
3 years ago
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