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S_A_V [24]
4 years ago
8

Solve please I’m desperate

Mathematics
2 answers:
NikAS [45]4 years ago
6 0

Answer:

-2

Step-by-step explanation:

\sqrt{16+9}-(\sqrt{16}+\sqrt{9})

We are going to use PE(MD)(AS) to evaluate.

We are going to do the operations in the grouping symbol.

A grouping symbol is anything that operations grouped together in it like that first square root.

\sqrt{25}-(\sqrt{16}+\sqrt{9})

We can also perform the operation inside the other grouping symbols ().

But it has addition and square roots. The square roots come first then the addition.

Now the square roots can be kind of though as exponents and actually you could think of something like \sqrt{9} as 9^{\frac{1}{2}}.

Anyways, this gives us:

\sqrt{25}-(4+3)  

Now the addition inside the ( ):

\sqrt{25}-(7)

Now we have square root and subtract left.  The square root part will come first and then the subtract:

5-7

-2

Ad libitum [116K]4 years ago
5 0
16 and 9 must be added as they are under the same root. We get 25 which is a perfect square. Next, 16 and 9 in parentheses are under different roots so they can’t be added. So, get the square root from 16 and 9 separately. This way we get 5-(4+3). Now, combine 5-(7). Next, multiply by - sign. This gives us a negative number. So, we get 5-7. Which equals -2. See the attachment for solving.

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Answer:

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What do i write for this and what do i draw??
Klio2033 [76]

The centre of the rotation is A

The line of the reflection is Y

Vertex A of the triangle ABC when rotated by 90° counterclockwise about the origin,

Rule to be followed,

A(x, y) → P(-y, x)

Therefore, A(1, 1) → P(-1, 1)

Similarly, B(3, 2) → Q(-2, 3)

C(2, 5) → R(-5, 2)

Triangle given in second quadrant will be the triangle PQR.

If the point P of triangle PQR is reflected across a line y = x,

Rule to be followed,

P(x, y) → X(y, x)

P(-1, 1) → X(1, -1)

Similarly, Q(-2, 3) → Y(3, -2)

R(-5, 2) → Z(2, -5)

Therefore, triangle given in the fourth quadrant is triangle XYZ.

Learn more about reflection

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6 0
2 years ago
Solve 5c 4 = -26 <br> a. 6 <br> b. -6 <br> c. -4.4 <br> d. 3
lakkis [162]
Simplifying 5C + -4 + -2C + 1 = 8C + 2 Reorder the terms: -4 + 1 + 5C + -2C = 8C + 2 Combine like terms: -4 + 1 = -3 -3 + 5C + -2C = 8C + 2 Combine like terms: 5C + -2C = 3C -3 + 3C = 8C + 2 Reorder the terms: -3 + 3C = 2 + 8C Solving -3 + 3C = 2 + 8C Solving for variable 'C'. Move all terms containing C to the left, all other terms to the right. Add '-8C' to each side of the equation. -3 + 3C + -8C = 2 + 8C + -8C Combine like terms: 3C + -8C = -5C<span>-3 + -5C = 2 + 8C + -8C Combine like terms: 8C + -8C = 0 -3 + -5C = 2 + 0 -3 + -5C = 2 Add '3' to each side of the equation. -3 + 3 + -5C = 2 + 3 Combine like terms: -3 + 3 = 0 0 + -5C = 2 + 3 -5C = 2 + 3 Combine like terms: 2 + 3 = 5 -5C = 5 Divide each side by '-5'. C = -1 Simplifying C = -1</span>
6 0
3 years ago
Kite EFGH is inscribed in a rectangle such that F and H are midpoints and EG is parallel to the side of the rectangle.
aivan3 [116]

Answer:

"The area of EFGH is always One-half of the area of the rectangle."

Step-by-step explanation:

<em>Graph is attached.</em>

<em />

The kite consists of 2 triangle, EFG and EHG.

The area of EFG:

\frac{1}{2}*EG*h

where h is the height from F to EG

The area of EHG:

\frac{1}{2}*EG*h_1

where h_1 is the height from H to EG

We also know that h + h _1 is the width of the rectangle and EG is the length of the rectangle

Thus,

Area of Kite = \frac{1}{2}*EG*h + \frac{1}{2}*EG*h_1 = RectangleLength*RectangleWidth

Also, Area of rectangle is rectangle length * rectangle width.

Thus, area of kite is always half of that of rectangle, the third choice is right.

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