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weqwewe [10]
3 years ago
15

Besties help me rn

Mathematics
1 answer:
olga55 [171]3 years ago
7 0

Answer:

(x+5)-(2x-3)

or simplified

8-x or -x+8

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Is 12.54 greater than 12.539?
Damm [24]

Answer:

yes

Step-by-step explanation:

12.54

>

12.53

reason why is because of the 4>5

8 0
3 years ago
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Distribute and combine like terms: −4(7r + 7)
Amiraneli [1.4K]

Answer:

r=1

Step-by-step explanation:

-4(7r+7)

-28r-28

r=1

6 0
3 years ago
Which set of data would be best modeled by a quadratic function?
katen-ka-za [31]
Graph B
Quadratic functions tend to portray a parabola (either an U shape or an opposite down U shape)
5 0
3 years ago
A "child has six blocks, three of which are red and three of which are green". How many patterns can shemake by placing them all
makvit [3.9K]

Answer: There are 20 ways and 1680 ways respectively.

Step-by-step explanation:

Since we have given that

Total number of blocks = 6

Number of red blocks = 3

Number of green blocks = 3

So, Number of patterns she can make by placing them all in a line is given by

\dfrac{6!}{3!\times 3!}\\\\=20

If there are 3 white blocks

so, total number of white blocks becomes 9

So, Number of total pattern she can make by placing all nine blocks in a line is given by

\dfrac{9!}{3!\times 3!\times 3!}\\\\=1680\ ways

Hence, there are 20 ways and 1680 ways respectively.

6 0
3 years ago
CAN SONEONE PLEASE HELP ME WITH MY MATH PLEASEEEE!!!!!​
Marina86 [1]

Answer:

√190

Step-by-step explanation:

In the figure , there are 2 right angled triangles with a common perpendicular & both the triangles combine to form a new right angled triangle.

Let the triangle with 9 as base be T¹ & Let the triangle with base 10 be T². Let the triangle formed by T¹ & T² be T³.

In T² ,

Hypotenuse = y

Base = 10

According to Pythagorean Theorem ,

(Hypotenuse)² = (Base)² + (Perpendicular)²

Hence, (Perpendicular)² = y^2 - 10^2 = y^2 - 100

In T¹ ,

Perpendicular = \sqrt{y^2 - 100}   (∵ Both T¹ & T² have common perpendicular)

⇒(Perpendicular)² = y^2 - 100

Base = 9

⇒ (Base)² = 9²

Hypotenuse =

Using Pythagorean Theorem ,

(Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ (Hypotenuse)² = y^2 - 100 + 9^2 .............................................eqn.2

Now in T³ ,

Base = y

⇒ (Base)² = y²

Perpendicular = \sqrt{(y^2 - 100) + 9^2} (∵Perpendicular of T³ = Hypotenuse of T²)

⇒ (Perpendicular)² = (\sqrt{(y^2 - 100) + 9^2})^2= (y^2 - 100) + 81 = y^2 - 19

Hypotenuse = 9 + 10 = 19

Using Pythagorean Theorem ,

(Hypotenuse)² = (Perpendicular)² + (Base)²

=> 19^2 = y^2 - 19 + y^2\\\\=> 2y^2 = 19^2 + 19 = 19(19 + 1) = 19*20\\\\=> y^2 = \frac{19*20}{2} = 19*10 = 190\\ \\=> y =\sqrt{190}

3 0
3 years ago
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