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vlada-n [284]
3 years ago
14

The standard equation of a circle with center (h.k) and radius r is (% - hy * (y - Ky = p. Which

Mathematics
1 answer:
zlopas [31]3 years ago
3 0

Answer:

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What is the value of |y|+4- y* x if x = 7 and y = -2?
irga5000 [103]

The given expression is

|y|+4-y\cdot x

Let's replace x = 7 and y = -2.

\begin{gathered} |-2|+4-(-2)\cdot7 \\ 2+4+14 \\ 20 \end{gathered}<h2>Therefore, the value of the expression is 20.</h2>
8 0
1 year ago
Solve each compound inequality. Graph your solutions.<br> 5 + m &gt; 4 or 7m&lt; -35
Yuliya22 [10]

Answer:

Solving the inequality 5 + m > 4\: or\: 7m< -35 we get \mathbf{m>-1\:or\: \:m

The graph of the inequality is shown in figure attached.

Step-by-step explanation:

We need to Solve each compound inequality 5 + m > 4\: or\: 7m< -35

Solving the inequalities

For solving compound inequalities we solve both inequalities simultaneously and find the value of m.

For solving 5+m > 4, we subtract 5 from both sides.

While solving 7m < -35, we divide both sides by 7 to get value of m

Applying these functions we get:

5 + m > 4\: or\: 7m< -35\\m>4-5\:or\:m-1\:or\:m

So, solving the inequality 5 + m > 4\: or\: 7m< -35 we get \mathbf{m>-1\:or\: \:m

The graph of the inequality is shown in figure attached.

5 0
2 years ago
Let T:V→W be a linear transformation from a vector space V into a vector space W. Prove that the range of T is a subspace of W.
9966 [12]

Answer:

The range of T is a subspace of W.

Step-by-step explanation:

we have T:V→W

This is a linear transformation from V to W

we are required to prove that the range of T is a subspace of W

0 is a vector in range , u and v are two vectors in range T

T = T(V) = {T(v)║v∈V}

{w∈W≡v∈V such that T(w) = V}

T(0) = T(0ⁿ)

0 is  Zero in V

0ⁿ  is zero vector in W

T(V) is not an empty subset of W

w₁, w₂   ∈ T(v)

(v₁, v₂ ∈V)

from here we have that

T(v₁) = w₁

T(v₂) = w₂

t(v₁) + t(v₂) = w₁+w₂

v₁,v₂∈V

v₁+v₂∈V

with a scalar ∝

T(∝v) = ∝T(v)

such that

T(∝v) ∈T(v)

so we have that T(v) is a subspace of W. The range of T is a subspace of W.

8 0
3 years ago
HELP PICTURE IS SHOWN
Sergeu [11.5K]
You will need option A
AB = PQ
8 0
3 years ago
Read 2 more answers
Solve 11b^2-9=-68 using the square root method
Vladimir79 [104]

Answer:

b=i*\sqrt{\frac{59}{11} } or -i*\sqrt{\frac{59}{11} }

Step-by-step explanation:

11b^2-9=-68

11b^2=-59

b^2=-59/11

b=i*\sqrt{\frac{59}{11} } or -i*\sqrt{\frac{59}{11} }

"i" in this case is an imaginary number, equal to \sqrt{-1\\}

if you haven't learned about these yet, something is wrong with the question

3 0
3 years ago
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