Ask question

Ask question

All categories

3 years ago

15

9) Shoe Box (L = 36 cm W= 20 cm H= 16 cm)

1 answer:

uysha [10]3 years ago

8 0

Answer:

zjfxkydkgdkydkydkyxyiditdigd

You might be interested in

Log5(x-1)+log5(x+3)-1=0

jonny [76]

Answer:

there are no real solutions

Step-by-step explanation:

$\log_5(x-1)+\log_5(x+3)-1=0\\\\\log_5(x-1)+\log_5(x+3)=1$

there is a rule that says

$\log_b(a)+\log_b(c)=\log_b(ac)$

so we have

$\log_5(x-1)+\log_5(x+3)=\log_5((x-1)(x+3))=1$

and we have the definition

$\log_a(b)=c\\\\a^{c} =b$

so we have

$5^{1} = (x-1)(x+3)\\\\5=x^{2} -1x+3x-3\\\\5=x^{2} +2x-3\\\\0=x^{2} +2x-3-5\\\\0=x^{2} +2x-8\\\\$

and using the quadratic formula we get that

there are no real solutions

3 0

4 years ago

Read 2 more answers

Zhen is an audiobook narrator. The table gives some completed audio lengths and the amount of time Zhen spends researching, reco

mestny [16]

Answer:

c. Zhen spends 6 1/2 hours working per audiobook.

Step-by-step explanation:

The table below gives some completed audio lengths and the amount of time Zhen spends on the audiobook.

$\left|\begin{array}{c|c}$Completed length of audiobook (hours) &$Working time (hours)\\---------&--------\\1&6\frac{1}{2} \\\\3\frac{1}{2}&22\frac{3}{4}\\\\7&45\frac{1}{2}\\\\10&65\end{array}\right|$

We observe that from the table, in 6.5 hours, Zhen completes 1 audiobook.

In fact, across the table:

$\dfrac{6\frac{1}{2}}{1}=\dfrac{22\frac{3}{4}}{3\frac{1}{2}}=\dfrac{65}{10}=6\dfrac{1}{2} $ hour per audiobook$

Therefore, in this situation, 6.5 hours mean that Zhen spends 6.5 hours working per audiobook.

The correct option is C.

5 0

3 years ago

Read 2 more answers

svetoff [14.1K]

Answer:

(a) Rule: Multiply the input (x) by 2 and add 5 to the result of the product

(b) $y =2x +5$

Step-by-step explanation:

Given

The attached table

Solving (a): The rule of the table in words

First, we calculate the slope (m)

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Using:

$(x_1,y_1) = (2,9)$

$(x_1,y_1) = (10,25)$

So, we have:

$m = \frac{25 - 9}{10 - 2}$

$m = \frac{16}{8}$

$m=2$

The equation is calculated using:

$y - y_1 =m(x - x_1)$

Using:

$(x_1,y_1) = (6,17)$

We have:

$y - 17 =2(x - 6)$

$y - 17 =2x - 12$

Make y the subject

$y =2x - 12+17$

$y =2x +5$

Hence, the rule is:

Multiply the input (x) by 2 and add 5 to the result of the product

(b): The rule, in algebra

As solved in (a), the rule is:

$y =2x +5$

8 0

3 years ago

The mean mark of a group of 10 boys is 58. If the mean of 7 of them is 61, what is the mean of the remaining 3 boys

Nesterboy [21]

Answer:

The mean of the remaining 3 boys is 51.

Step-by-step explanation:

The mean mark of the entire group ( $p_{10}$ ), dimensionless, is:

$p_{10} = \frac{1}{10}\cdot \Sigma_{i = 1}^{10} x_{i}$ (Eq. 1)

( $p_{10} = 58$ )

$\frac{1}{10}\cdot \Sigma_{i=1}^{10} x_{i} = 58$ (Eq. 1b)

Where $x_{i}$ is the mark of the i-th boy, dimensionless.

In addition, we know the following mean marks from statement:

$p_{7} = \frac{1}{7} \cdot \Sigma_{i = 1}^{7} x_{i}$ (Eq. 2)

( $p_{7} = 61$ )

$\frac{1}{7}\cdot \Sigma_{i=1}^{7} x_{i} = 61$ (Eq. 2b)

$p_{3} = \frac{1}{3}\cdot \Sigma_{i=8}^{10}x_{i}$ (Eq. 3)

Where:

$p_{7}$ - Mean mark of the first 7 boys, dimensionless.

$p_{3}$ - Mean mark of the remaining 3 boys, dimensionless.

By applying sum properties in (Eq. 1b) and using (Eq. 2b) and (Eq. 3), we obtain the mean of the remaining 3 boys:

$\frac{1}{10}\cdot [\Sigma_{i = 1}^{7}x_{i}+\Sigma_{i=8}^{10}x_{i}] = 58$

$\frac{1}{10}\cdot [7\cdot (61)+3\cdot p_{3}] = 58$

$7\cdot (61) + 3\cdot p_{3} = 580$

$3\cdot p_{3} = 153$

$p_{3} = \frac{153}{3}$

$p_{3} = 51$

The mean of the remaining 3 boys is 51.

7 0

3 years ago

V=4/3nr^3<br>r=7<br>solve for pie

Aleks04 [339]

1372/3n I hope this helped :)

3 0

4 years ago