All categories

3 years ago

Please please help meeee

Mathematics

2 answers:

Sladkaya [172]3 years ago

8 0

Answer:

Step-by-step explanation:

10/a = 15/24

Cross multiply.

15 × a = 10 × 24

15a = 240

Divide both sides with 15.

a = 240/15

a = 16

larisa [96]3 years ago

3 0

Answer:

a = 16

Step-by-step explanation:

=> $\frac{10}{a} = \frac{15}{24}$

Cross Multiplying

=> 10 * 24 = a * 15

=> 15a = 240

Dividing both sides by 15

=> a = 16

You might be interested in

Let the random variable x be equally likely to assume any of the values 1/10, 1/5, or 3/10. determine the mean and variance of x

Keith_Richards [23]

Since the values are equally likely and equally spaced, their mean is the middle one.
the mean is 1/5

The variance is the average of the square of the deviations from the mean.
σ² = (1/3)((-1/10)² + 0² + (1/10)²) = 2/300
the variance is 2/300

6 0

3 years ago

What is the measure of ∠ADC in quadrilateral ABCD? °

anastassius [24]

115 is the correct answer.
I saw the diagram earlier :)
Hope this helps, have a great day, and God bless.
Brainliest is always appreciated :)

7 0

4 years ago

looks out from the crown of the Statue of Liberty, approximately 250 ft above ground. The tourist sees a ship coming into the ha

Zigmanuir [339]

Answer:

The distance from the base of the statue to the ship is 771.60 ft

Step-by-step explanation:

Refer the attached figure

Height of statue AB= 250 feet

The tourist sees a ship coming into the harbor and measures the angle of depression as 18°.

So, ∠ACB = 18°

We are supposed to find the distance from the base of the statue to the ship i.e. BC

In ΔABC

$Tan \theta = \frac{Perpendicular}{Base}$

$Tan 18^{\circ}=\frac{AB}{BC}$

$Tan 18^{\circ}=\frac{250}{BC}$

$BC=\frac{250}{Tan 18^{\circ}}$

$BC=\frac{250}{0.324}$

BC=771.60 ft

Hence the distance from the base of the statue to the ship is 771.60 ft

3 0

3 years ago

What is (3x+5)(x^2+6x+11)

serg [7]

Answer:

$\large\boxed{3x^3+23x^2+63x+55}$

Step-by-step explanation:

$\text{Use FOIL:}\\\\(a+b)(c+d)=ac+ad+bc+bd$

$(3x+5)(x^2+6x+11)\\\\=(3x)(x^2)+(3x)(6x)+(3x)(11)+(5)(x^2)+(5)(6x)+(5)(11)\\\\=3x^3+18x^2+33x+5x^2+30x+55\qquad\text{combine like terms}\\\\=3x^3+(18x^2+5x^3)+(33x+30x)+55\\\\=3x^3+23x^2+63x+55$

3 0

4 years ago

I need helppppppp. whats 1/5x30

Vadim26 [7]

Answer:

Step-by-step

You're welcome!

3 0

3 years ago