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pychu [463]
2 years ago
11

3. Solve 2log4y - log4 (5y - 12) = 1/2​

Mathematics
1 answer:
Ilia_Sergeevich [38]2 years ago
4 0

Answer:

y =  4  or y = 6

Step-by-step explanation:

2log4y - log4 (5y - 12) = 1/2

​2log_4(y) - log_4(5y-12) = log_4(2)           apply law of logarithms

log_4(y^2) + log_4(1/(5y-12)) = log_4(/2)    apply law of logarithms

log_4(y^2/(5y-12)) = log_4(2)                     remove logarithm

y^2/(5y-12) = 2                                            cross multiply

y^2 = 10y-24                                                  rearrange and factor

y^2 - 10y + 24 = 0

(y-4)(y-6) = 0

y= 4 or y=6

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Darcie wants to crochet a minimum of 3 blankets. Darcie crochets at a rate of 1/15 of a blanket a day. She has 60 days until she
NNADVOKAT [17]

Answer:

The inequality to determine the number of days, s, Darcie can skip crocheting and still meet her goal can be given as:

s\leq 15

Step-by-step explanation:

The complete question is:

Darcie wants to crochet a minimum of 3 blankets to donate to a homeless shelter. Darcie crochets at a rate of 1/15 of a blanket per day. She has 60 days until when she wants to donate the blankets, but she also wants to skip crocheting some days so she can volunteer in other ways. Write an inequality to determine the number of days, s, Darcie can skip crocheting and still meet her goal.

Solution:

Given:

Darcie wants to crochet a minimum of 3 blankets to donate.

Rate at which she crochet = \frac{1}{15} of a blanket per day.

Maximum number of days she has = 60.

To find the number of days Dancie can skip out of 60 days and still reach her goal.

Let s represent the number of days she can skip.

Number of days left to crochet = (60-s) days

At rate of  \frac{1}{15} of a blanket per day, number of blankets Dancie can corchet in (60-s) days can be given as :

⇒ \frac{1}{15}(60-s)

Simplifying using distribution.

⇒ (\frac{1}{15}.60)-(\frac{1}{15}.s)

⇒ 4-\frac{s}{15}

Dancie needs to crochet a minimum of 3 blankets to reach her goal.

Thus, the inequality can be given as:

4-\frac{s}{15}\geq 3

Solving the inequality for s

Subtracting 4 both sides.

4-4-\frac{s}{15}\geq 3-4

-\frac{s}{15}\geq -1

Multiplying both sides by -15.

-15(-\frac{s}{15})\leq -15(-1) [On multiplying by a negative number the sign of inequality reverse]

∴ s\leq 15

Thus, Dancie can skip a maximum of 15 days.

5 0
2 years ago
The Goodsmell perfume producing company has a new line of perfume and is
LenKa [72]

Given:

It is given that surface area must be less than 150 cm².

Solution:

The Maximum Volume With Total Surface Area Less  than 150 cm² is shown in the table.

From the table, it can be concluded that for r=3.00 cm and h=4.95 cm the surface area will be less than 150 cm²  and the volume will be the maximum.

S=2\pi rh+2\pi r^2\\S=2\pi (3)(7.95)+2\pi3^2\\S=93.3+56.5\\S=149.8 \text{ cm}^2

Calculate the volume.

V=\pi r^2h\\V=\pi(3)^2(4.95)\\V=139.96

Hence, the required dimensions are r=3.00 cm and h=4.95 cm.

8 0
2 years ago
The experimental probability that Kevin will catch a fly ball is equal to 7/8. About what percent of the time will Kevin catch a
Paladinen [302]

Answer:

87.5% percent chance.

Trust me it's right.

6 0
2 years ago
A machine is supposed to mix peanuts, hazelnuts, cashews, and pecans in the ratio 5:2:2:1. A can containing 500 of these mixed n
luda_lava [24]

Answer:

the machine is mixing the nuts are not  in the ratio 5:2:2:1.

Step-by-step explanation:

Given that a machine is supposed to mix peanuts, hazelnuts, cashews, and pecans in the ratio 5:2:2:1.

A can containing 500 of these mixed nuts was found to have 269 peanuts, 112 hazelnuts, 74 cashews, and 45 pecans.

Create hypotheses as

H0: Mixture is as per the ratio 5:2:2:1

Ha: Mixture is not as per the ratio

(Two tailed chi square test)

Expected values as per ratio are calculated as 5/10 of 500 and so on

Exp        250      100    100       50        500

Obs       269      112       74       45         500

O-E          19        -12      -26       -5           0

Chi          1.343   1.286  9.135   0.556   12.318

square

df = 3

p value = 0.00637

Since p value < alpha, we reject H0

i.e. ratio is not as per the given

4 0
2 years ago
Find all x in set of real numbers R Superscript 4 that are mapped into the zero vector by the transformation Bold x maps to Uppe
sukhopar [10]

Answer:

 x_3 = \left[\begin{array}{c}4&3&1\\0\end{array}\right]

Step-by-step explanation:

According to the given situation, The computation of all x in a set of a real number is shown below:

First we have to determine the \bar x so that A \bar x = 0

\left[\begin{array}{cccc}1&-3&5&-5\\0&1&-3&5\\2&-4&4&-4\end{array}\right]

Now the augmented matrix is

\left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\2&-4&4&-4\ |\ 0\end{array}\right]

After this, we decrease this to reduce the formation of the row echelon

R_3 = R_3 -2R_1 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\0&2&-6&6\ |\ 0\end{array}\right]

R_3 = R_3 -2R_2 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\0&0&0&-4\ |\ 0\end{array}\right]

R_2 = 4R_2 +5R_3 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&4&-12&0\ |\ 0\\0&0&0&-4\ |\ 0\end{array}\right]

R_2 = \frac{R_2}{4},  R_3 = \frac{R_3}{-4}  \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&1\ |\ 0\end{array}\right]

R_1 = R_1 +3 R_2 \rightarrow \left[\begin{array}{cccc}1&0&-4&-5\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&-1\ |\ 0\end{array}\right]

R_1 = R_1 +5 R_3 \rightarrow \left[\begin{array}{cccc}1&0&-4&0\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&-1\ |\ 0\end{array}\right]

= x_1 - 4x_3 = 0\\\\x_1 = 4x_3\\\\x_2 - 3x_3 = 0\\\\ x_2 = 3x_3\\\\x_4 = 0

x = \left[\begin{array}{c}4x_3&3x_3&x_3\\0\end{array}\right] \\\\ x_3 = \left[\begin{array}{c}4&3&1\\0\end{array}\right]

By applying the above matrix, we can easily reach an answer

5 0
1 year ago
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