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3 years ago

-3r+15≥4(r-2) please help me

2 answers:

5 0

-3r + 15 ≥ 4(r - 2)
-3r + 15 ≥ 4(r) - 4(2)
-3r + 15 ≥ 4r - 8
+ 3r    + 3r
   15 ≥ 7r - 8
      + 8    + 8
       23 ≥ 7r
      7    7
      3²/₇ ≥ r
      r ≤ 3²/₇

Solution Set: x ∈ (-∞, 3²/₇]

adell [148]3 years ago

5 0

−3r+15≥4r−8

Subtract 4r from both sides.

−3r+15−4r≥4r−8−4r

−7r+15≥−8

Subtract 15 from both sides.

−7r+15−15≥−8−15

−7r≥−23

Divide both sides by -7.

−7r/−7 ≥ −23/−7

r ≤ 23/7

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Find the modulus of the complex number 6-2i

Papessa [141]

Answer:

The modulus of the complex number 6-2i is:

$|z|\:=2\sqrt{10}$

Step-by-step explanation:

Given the number

$6-2i$

We know that

$z = x + iy$

where x and y are real and $\sqrt{-1}=i$

The modulus or absolute value of z is:

$|z|\:=\sqrt{x^2+y^2}$

Therefore, the modulus of $6-2i$ will be:

$z=6-2i$

$z=6+(-2)i$

$|z|\:=\sqrt{x^2+y^2}$

$|z|\:=\sqrt{6^2+\left(-2\right)^2}$

$=\sqrt{6^2+2^2}$

$=\sqrt{36+4}$

$=\sqrt{40}$

$=\sqrt{2^2}\sqrt{2\cdot \:5}$

$=2\sqrt{2\cdot \:5}$

$=2\sqrt{10}$

Therefore, the modulus of the complex number 6-2i is:

$|z|\:=2\sqrt{10}$

3 0

3 years ago

estimate 0.037854921 to the nearest hundredth express your answer as a single digit times a power of 10

Vilka [71]

To solve this problem you must apply the proccedure shown below:

1. Let's round the value to the nearest hundredth. As you can see, the digit 8 is in the thousandths place and is greater than 5, therefore, you must round up to 0.038.

2. Now express the value as a single digit times a power of 10, as following:

$38$ x $10^{-3}$