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3 years ago

A point in quadrant IV is reflected across the x-axis. The new point is located in Quadrant..?

2 answers:

8 0

A point in quadrant IV is reflected across the x-axis. The new point is located in Quadrant..?
A) I
B) II
C) III
D) IV

It’s C) III

suter [353]3 years ago

4 0

Answer

It's D) IV

Step-by-step explanation:

Just look at a graph.

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lidiya [134]

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3 years ago

Read 2 more answers

3-10-(-15)<br> Show and explain work

Genrish500 [490]

Answer:

3-10-(15) equals 8.

Step-by-step explanation:

3-10 equals -7. Then -(-15) becomes a positive because two negatives make a positives producing +15 so it becomes -7+15 which finally becomes 8.

3 0

4 years ago

For what value of "a" will the product shown below have a purely imaginary value? (Show work please)

MakcuM [25]

If you expand out the brackets you get this,

(4+5i)(a+2i) = 4a + (5a)i + 8i - 10

The -10 comes from 5i * 2i.
Squaring i becomes -1.

Let's group the real stuff together,
and imaginary separately,

(4a - 10) + (5a + 8)i

For this to be purely imaginary,
the real part needs to be zero.

Therefore 4a - 10 = 0

Solve for a.

5 0

3 years ago

The equation

ra1l [238]

<h2>Explanation:</h2>

Let's use a trial and improvement method to find this solution.

Step 1. Let's choose x = 8.5

Substituting into the equation:

$(8.5)^2+ 3(8.5) - 94=0? \\ \\ 72.25+25.5-94=0? \\ \\ 3.75\neq 0$

Step 2. Let's choose x = 8.4

Substituting into the equation:

$(8.4)^2+ 3(8.4) - 94=0? \\ \\ 70.56+25.2-94=0? \\ \\ 1.76\neq 0$

Step 3. Let's choose x = 8.3

Substituting into the equation:

$(8.3)^2+ 3(8.3) - 94=0? \\ \\ 68.89+24.9-94=0? \\ \\ -0.21\neq 0$

Since the sign of the equation changes from positive to negative when evaluating from 8.4 to 8.3, then x = 8.3 seems to be a reasonable value. Finally, the solution to 1 decimal place is:

$x=8.3$

6 0

3 years ago

Help me I don’t get this.

Marizza181 [45]

Answer:

i think its 105

Step-by-step explanation:

5 0

3 years ago