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3 years ago

Please could someone help me?

Mathematics

2 answers:

evablogger [386]3 years ago

8 0

Yes I can boi I’m very smart very smart like mega mind

Sonja [21]3 years ago

3 0

Answer:

∠ XZY ≈ 23.6°

Step-by-step explanation:

Using the sine ratio in the right triangle

sin Z = $\frac{opposite}{hypotenuse}$ = $\frac{XY}{YZ}$ = $\frac{6}{15}$ , thus

Z = $sin^{-1}$ ( $\frac{6}{15}$ ) ≈ 23.6° ( to 1 dec. place )

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Estimate 5,738,055 × 32.

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The exact answer is 183,617,760. Rounding to the nearest convenient numbers, which would make it 5,750,000 x 30, the estimated answer would be 172,500,000

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4 years ago

There are 25 students who answered a survey about sports. Four-fifths of them like

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Answer: 20 students

Step-by-step explanation:

you do 25 divided by 4/5

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4 years ago

Read 2 more answers

i) Baraka bought ksh.20,000 worth of shares of a company each month. During the first three months, he bought the shares at a pr

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The average price paid by him for the shares after 3 months is ksh. 163.33

<h3>Average</h3>

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2 years ago

The number of research articles in a prominent journal written by researchers during 1983–2003 can be approximated by U(t) = 4.1

Nimfa-mama [501]

Answer:

Step-by-step explanation:

From the information given:

We are to find; an expression for the total no. of articles written since 1983

∴

The total no. of articles written since 1983 $=\int \limts ^t_0 U(t) dt$

$= \int ^t_0 \limits \dfrac{4.1 e^{0.6\ t}}{0.2+e^{0.6t}} \ dt$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{0.6 x} \bigg| \bigg]^t_0$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{0.6 x} \bigg| - \mathtt{In} \bigg|0.2+1\bigg|\bigg]$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | \dfrac{0.2 + e^{0.6 x}}{1.2} \bigg|\bigg]$

$=\dfrac{41}{6} \mathtt{In} \bigg | {0.2 + e^{0.6 x} \bigg|-\dfrac{41}{6} \mathtt{ In} \bigg | 1.2\bigg |$

$=6.833\times \mathtt{In} \bigg | {0.2 + e^{0.6 x} \bigg|-1.2458$

$=6.833\times \mathtt{In} \bigg | {0.2 + e^{0.6 x} \bigg|-1.25 \ \mathbf{thousand \ articles}$

Therefore, the total number of articles written since 1983 is $=6.833\times \mathtt{In} \bigg | {0.2 + e^{0.6 x} \bigg|-1.25 \ \mathbf{thousand \ articles}$

b. To find how many articles were being written from 1983 to 2003

i.e. t = 2003 - 1983 = 20

∴

Total articles written from 1983 to 2003 is $=\int \limts ^{20}_0 U(t) dt$

$= \int ^{20}_0 \limits \dfrac{4.1 e^{0.6\ t}}{0.2+e^{0.6t}} \ dt$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{0.6 x} \bigg| \bigg]^{20}_0$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{0.6 *20} \bigg| - \mathtt{In} \bigg|0.2+e^{0.6*0}\bigg|\bigg]$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{12} \bigg| - \mathtt{In} \bigg|0.2+e^0\bigg|\bigg]$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{12} \bigg| - \mathtt{In} \bigg|0.2+1\bigg|\bigg]$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{12} \bigg| - \mathtt{In} \bigg|1.2\bigg|\bigg]$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | \dfrac{0.2 + e^{12}}{1.2} \bigg|\bigg]$

= 80.75 thousand articles

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3 years ago

10,000 candy canes was divided into 50 boxes. How many candy canes were in each box?

kumpel [21]

Answer:

200

Step-by-step explanation:

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2 years ago