All categories

3 years ago

Help like I do not understand math at all

Mathematics

2 answers:

Fynjy0 [20]3 years ago

7 0

Parts A:There will be 57 rows
Part B: because 513 divided by 9 equals 57
Part C:equals 25 because if you divide 250 by 5 and then divide it by 2 u get 25

yKpoI14uk [10]3 years ago

5 0

Answer:

Step-by-step explanation:

You might be interested in

PLEASE SHOW WORK OR EXPLAIN

timofeeve [1]

Answer:

Step-by-step explanation:

You must use the trigonometric function.

In this case, it will be tangent.

$tangent=\dfrac{opposite}{adjacent}$

We have

$\theta=38^o,\ opposite=r,\ adjacent=20$

$\tan38^\approx0.7813$

Substitute:

$\tan38^o=\dfrac{r}{20}$

$\dfrac{r}{20}=0.7813$ <em>multiply both sides by 20</em>

$r=15.626$

Tree height is r + 5.

Therefore

$r+5=15.626+5=20.626\approx20.6\ ft$

7 0

3 years ago

How can you you tell when a pattern shows counting on by tens

astraxan [27]

You can ise repeated reasonings

6 0

3 years ago

What assumption have you made in answering part b

emmasim [6.3K]

You're assuming that this data set represents all of Mrs. Hassaan's pupils. Certain students could be missing from the histogram - for instance, there could be one or two students who missed the test, but for certain reasons will be taking a make-up exam. Their grade would not be recording in this data set, but they still contribute to the class size, and so would be unaccounted for here.

4 0

3 years ago

Given the following right triangle, find cose, sine, tane sececsce, and cote. Do not

IgorLugansk [536]

Step-by-step explanation:

Use the Pythagorean theorem:

$leg^2+leg^2=hypotenuse^2$

We have

$leg=4,\ hypotenuse=10$

Substitute:

$4^2+leg^2=10^2$

$16+leg^2=100$ <em>subtract 16 from both sides</em>

$leg^2=84\to leg=\sqrt{84}\\\\leg=\sqrt{(4)(21)}\\\\leg=\sqrt4\cdot\sqrt{21}\\\\leg=2\sqrt{21}$

$sine=\dfrac{opposite}{hypotenuse}\\\\cosine=\dfrac{adjacent}{hypotenuse}\\\\tangent=\dfrac{opposite}{adjacent}\\\\cotangent=\dfrac{adjacent}{opposite}\\\\secant=\dfrac{hypotenuse}{adjacent}\\\\cosecant=\dfrac{hypotenuse}{opposite}$

Substitute:

$hypotenuse=10,\ opposite=4,\ adjacent=2\sqrt{21}$

$\sin\theta=\dfrac{4}{10}=\dfrac{2}{5}\\\\\cos\theta=\dfrac{2\sqrt{21}}{10}=\dfrac{\sqrt{21}}{5}\\\\\tan\theta=\dfrac{4}{2\sqrt{21}}=\dfrac{2}{\sqrt{21}}\cdot\dfrac{\sqrt{21}}{\sqrt{21}}=\dfrac{2\sqrt{21}}{21}\\\\\cot\theta=\dfrac{2\sqrt{21}}{4}=\dfrac{\sqrt{21}}{2}\\\\\sec\theta=\dfrac{10}{2\sqrt{21}}=\dfrac{5}{\sqrt{21}}\cdot\dfrac{\sqrt{21}}{\sqrt{21}}=\dfrac{5\sqrt{21}}{21}\\\\\csc\theta=\dfrac{10}{4}=\dfrac{5}{2}$

5 0

4 years ago

What is the density of 72g over 24cm?

diamong [38]

Mass/volume 72g/24 cubic cm Is equal to 3G per cubic centimeter

5 0

3 years ago