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Snezhnost [94]
3 years ago
14

Is 619 good on an I-ready test?? It was on math/7th grade

Mathematics
2 answers:
Anestetic [448]3 years ago
8 0

Answer: If i'm not mistaking it is, but i haven't did I-ready in a while.

Irina-Kira [14]3 years ago
3 0

Answer: a what test?

Step-by-step explanation:

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How do I do this? Find the distance between the points given. Simplify all irrational answers.
belka [17]
The formula for finding the distance between points is √(x-x₁)²+(y-y₁)²
x=0, y=6, x₁=5, y₁=12
√(0-5)²+(6-12)²
√(-5)²+(-6)²
√25+36
√61 is irrational
7.81 points(rounded to nearest hundredth) is the distance between those two points.
8 0
4 years ago
Help me I’m dumb and stupid :)
wel

Answer:

The first digit of the quotient should be placed at the leftmost place of the places of the all the digits in the quotient.This is so from the basic rule of division.

Step-by-step explanation:

The quotient is given by,

[\frac {4,839}{15}] [where [x] is the greatest integer function on x]

= [322.6]

= 322

and the remainder is given by,

15 \times 0.6

= 9

So, the first digit of the quotient should be placed at the leftmost place of the places of the all the digits in the quotient and this is so from the very basic rule of division.

8 0
3 years ago
A woman who has recovered from a serious illness begins a diet regimen designed to get her back to a healthy weight. She current
WINSTONCH [101]

Answer:

a.) w = 103 * 1.08^t

b.) 3.5weeks

Step-by-step explanation:

If Her current weight is 103 pounds and she hopes to multiply her her weight each week by 1.08, then

her weight after 1 week = 103 * 1.08 = 103 * 1.08¹

Her weight after 2 weeks = [weight of week 1] * 1.08 = [103* 1.08] * 1.08 = 103 * 1.08²

Weight after 3 weeks= [weight of week 2] * 1.08 = [103 * 1.08 * 1.08] * 1.08 = 103 * 1.08³

Hence weight (W) after t weeks = 103 * 1.08^t

b.) If W = 135, Then

103 * 1.08^t = 135

1.08^t = 135/103

1.08^t = 1.31

Taking log of both sides,

log 1.08^t = log 1.31

t log 1.08 = log 1.32

t = log 1.32/log 1.08

t = 3.5 weeks.

Hence, it will take her 3½ weeks to get to 135pounds weight.

7 0
3 years ago
Find the center of mass of the wire that lies along the curve r and has density =4(1 sin4tcos4t)
dolphi86 [110]

The mass of the wire is found to be 40π√2 units.

<h3>How to find the mass?</h3>

To calculate the mass of the wire which runs along the curve r ( t ) with the density function δ=5.

The general formula is,

Mass = \int_a^b \delta\left|r^{\prime}(t)\right| d t

To find, we must differentiate this same given curve r ( t ) with respect to t to estimate |r'(t)|.

The given integration limits in this case are a = 0, b = 2π.

Now, as per the question;

The equation of the curve is given as;

r(t) = (4cost)i + (4sint)j + 4tk

Now, differentiate this same given curve r ( t ) with respect to t.

\begin{aligned}\left|r^{\prime}(t)\right| &=\sqrt{(-4 \sin t)^2+(4 \cos t)^2+4^2} \\&=\sqrt{16 \sin ^2 t+16 \cos ^2 t+16} \\&=\sqrt{16\left(\sin t^2+\cos ^2 t\right)+16}\end{aligned}

Further simplifying;

\begin{aligned}&=\sqrt{16(1)+16} \\&=\sqrt{16+16} \\&=\sqrt{32} \\\left|r^{\prime}(t)\right| &=4 \sqrt{2}\end{aligned}

Now, use integration to find the mass of the wire;

       \begin{aligned}&=\int_a^b \delta\left|r^{\prime}(t)\right| d t \\&=\int_0^{2 \pi} 54 \sqrt{2} d t \\&=20 \sqrt{2} \int_0^{2 \pi} d t \\&=20 \sqrt{2}[t]_0^{2 \pi} \\&=20 \sqrt{2}[2 \pi-0] \\&=40 \pi \sqrt{2}\end{aligned}

Therefore, the mass of the wire is estimated as 40π√2 units.

To know more about density function, here

brainly.com/question/27846146

#SPJ4

The complete question is-

Find the mass of the wire that lies along the curve r and has density δ.

r(t) = (4cost)i + (4sint)j + 4tk, 0≤t≤2π; δ=5

5 0
2 years ago
Don't go by my answers pls
Illusion [34]

Answer:

19)  ( \frac{-2}{5})^2 = \frac{4}{25}

20) -9^{2} = -81

21)  2^{4} = 16

Step-by-step explanation:

The only wrong ones are

19)  ( \frac{-2}{5})^2 = \frac{4}{25}

20) -9^{2} = -81 ... negative is not inside a parenthesis that's why it is still negative

21) you can add another way is 2^{4} = 16

Everything else looks good.  Great job!

4 0
3 years ago
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