All categories

3 years ago

Height of 12 and base of 5 find the hypotanues

1 answer:

3 0

Height = 12 cm
Base = 5 cm

Hypotenuse^2 = Height^2 + Base ^2
= 12^2 + 5^2
= 144+25
Hypotenuse^2= 169
Hypotenuse = square root 169
= 13 cm

Therefore, the hypotenuse is 13cm

You might be interested in

Y =1/4x + 3<br> Y = -1<br> Help me

kap26 [50]

Y = 7

Explication pas à pas:

x est -3 donc -3 et connectez le x.

Y = -1 / 3 (-3) +5.

Y = 1 + 5

Y = 7

7 0

3 years ago

Sweetums finds that the cost of making a certain candy bar .15 per bar and that the fixed rate is 600 per week if each bar sells

PilotLPTM [1.2K]

We know that:

Profit = Revenue - Cost

Let us say x number of candies are made per week.

Finding Cost per week:

Cost of making 1 bar = 0.15

Cost of making x bars = 0.15x

Fixed rate of making candies per week = 600

Total cost of making x candies per week = 600 + 0.15x

Now let us find Revenue:

Selling price of each bar = 1.50

Selling price of x bars = 1.50x

Now we have to find profit,

Profit = Revenue - Cost

In order to have profit Revenue - Cost >0

So plugging values of revenue and cost to get number of candies,

$1.50x-(0.15x+600)>0$

$1.50x-0.15x-600>0$

$1.35x-600>0$

$1.35x>600$

x>444.44

Rounding off

x>444

Answer: The company must sell greater than 444 candies in order to make profit.

7 0

2 years ago

How do we write this in scientific notation?

marusya05 [52]

Answer:

The answer is Sodium chloride.

7 0

2 years ago

Read 2 more answers

Reduce fraction: a^3+a^2b/5a times 25/3b+3a

Margaret [11]

ANSWER

$\frac{5a}{3}$

EXPLANATION

The given fractions are:

$\frac{{a}^{3} + {a}^{2} b}{5a} \times \frac{25}{3b + 3a}$

We factor to obtain:

$\frac{{a}^{2}(a + b)}{5a} \times \frac{25}{3(a + b)}$

We cancel the common factors to get:

$\frac{{a}(1)}{1} \times \frac{5}{3(1)}$

We multiply the numerators and also multiply the denominators to get:

$\frac{5a}{3}$

Therefore the two fractions simplifies to $\frac{5a}{3}$

3 0

3 years ago

In a missile-testing program, one random variable of interest is the distance between the point at which the missile lands and t

Arisa [49]

Answer:

Step-by-step explanation:

From the given data

we observed that the missile testing program

Y1 and Y2 are variable, they are also independent

We are aware that

$(Y_1)^2 and (Y_2)^2$ have $x^2$ distribution with 1 degree of freedom

and $V=(Y_1^2)+(Y_2)^2$ has x^2 with 2 degree of freedom

$F_v(v)=\frac{e^{-\frac{v}{2}}}2$

Since we have to find the density formula

$U=\sqrt{V}$

We use method of transformation

$h(V)=\sqrt{U}\\\\=U$

There inverse function is $h^-^1(U)=U^2$

We derivate the fuction above with respect to u

$\frac{d}{du} (h^-^1(u))=\frac{d}{du} (u^2)\\\\=2u^2^-^1\\\\=2u$

Therefore,

$F_v(u)=F_v(h-^1)(u)\frac{dh^-^1}{du} \\\\=\frac{e^-\frac{u^-^}{2} }{2} (2u)\\\\=e^-{\frac{u^2}{2} }U$

8 0

3 years ago