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sertanlavr [38]
3 years ago
10

Work out the answers of the following:

Mathematics
2 answers:
schepotkina [342]3 years ago
7 0

Answer:

1)-5

2)-15

3)1

4)-11

5)-70

if negative use () for example

(-2)-7+4

kirill [66]3 years ago
6 0

Answer:

1. -5

2. - 15

3. 1

4. - 11

5. -70

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5. What is the equation of the tangent line to the circle x^2+y^2=1 through the point (6,0)?
notka56 [123]

Answer:

There is no tangent line of the given circle at (6, 0).

Step-by-step explanation:

Given equation of the circle,

x^2 + y^2 = 1

∵ equation of a circle is (x-h)^2 +(y-k)^2 = r^2,

Where, (h, k) is the center of the circle and r is the radius,

By comparing,

Center of the given circle = (0, 0),

Radius of the circle = 1 unit

Now, check whether point (6, 0) lie on the circle,

if x = 6, 6^2 + y^2 = 1

36 + y^2 = 1

y^2 = 1- 36

y= i\sqrt{35}\neq 0

i.e., (6, 0) does not lie on the circle,

Hence, there is no tangent line of the given circle at (6, 0).

3 0
3 years ago
Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the
S_A_V [24]

Answer:

8^2 = b^2 + 4^2

Step-by-step explanation:

5 0
2 years ago
Help me plzzzzzzzzzzz
Inga [223]

Answer:

A = 3

B = -5

C = 2

Step-by-step explanation:

The way to format a quadratic equation is: ax^2 + bx + c, so the first step to solving this is to format it in the right way, where the x^2 comes first, the x second, and the number alone.

After formatting, your equation should look like this: 3x^2 - 5x + 2

From here, you can see the the <em>a </em> is 3, the <em>b</em> is -5, and the <em>c </em>is 2

7 0
3 years ago
27 • a • a • a as an expression?
Alekssandra [29.7K]

the expression is 27a^3   or (3a)^3

5 0
3 years ago
Read 2 more answers
Solve each of the following for the indicated variable.a. 9 − 6 = 54 for yb. = 2 + 2 for l
marta [7]

Answer:

\begin{gathered} y=\frac{9}{6}x-9 \\ l=\frac{P}{2}-w \end{gathered}

Step-by-step explanation:

To solve the following equations, we need to isolate the variable asked using inverse operations to solve equations.

Remember that addition and subtraction are inverse operations, as multiplication and division.

Therefore,

\begin{gathered} a\text{. Solve for y} \\ 6y=9x-54 \\ y=\frac{9}{6}x-\frac{54}{6} \\ y=\frac{9}{6}x-9 \end{gathered}\begin{gathered} b\text{. Solve for l} \\ P=2l+2w \\ 2l=P-2w \\ l=\frac{P}{2}-\frac{2}{2}w \\ l=\frac{P}{2}-w \end{gathered}

4 0
1 year ago
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