All categories

3 years ago

Can someone help me on 13 Plz

Mathematics

2 answers:

Ad libitum [116K]3 years ago

7 0

10(3) + 3 = 33 times in 2 minutes
10(120) + 3 = 1203 times in 2 hours

Phoenix [80]3 years ago

6 0

What we know :
After m minutes, and adult blinks about 10m+3 times.
1 hour= 60 minutes

What we need to find:
About how many times does an adult blink in 3 minutes? In 2hours?

m=3
f (m)=10m+3
f (3)=10 (3)+3
=30+3
=33
An adult blinks 33 times in 3 minutes

2hours=2×60=120minutes

f (120)=10 (120)+3
=1200+3
=1203

An adult blinks 1203 times in 2 hours.

You might be interested in

Write the rational number -1, in the form , where a and b are<br> integers.

iogann1982 [59]

Answer:

2nd option

Step-by-step explanation:

- 5 = $\frac{-5}{1}$ ← in the form $\frac{a}{b}$

3 0

3 years ago

Use two different methods to find an explain the formula for the area of a trapezoid that has parallel sides of length a and B a

evablogger [386]

Answer:

Formula of Trapezoid:

A = (a + b) × h / 2

The formula can be derived in different ways. for now, we have discussed two ways:

1. By using the formula of a triangle

2. By dividing into different sections

Step-by-step explanation:

1. By using the formula of a triangle

One of the ways to explain a formula for an area of a trapezoid using a formula for a triangle can be as follows.

Assume a trapezoid PQRS with lower base SR and upper base PQ (they are parallel) and sides PS and QR.

The image is attached below.

Connect vertices P and R with a diagonal.

Consider triangle ΔPQR as having a base PQ and an altitude from vertex R down to point M on base PQ (RM⊥PQ).

Its area is

S1= $\frac{1}{2} *PQ*RM$

Consider triangle ΔPRS as having a base SR and an altitude from vertex P up to point N on-base SR (PN⊥SR).

Its area is

S2= $\frac{1}{2} *SR*PN$

Altitudes RM and PN are equal and constitute the distance between two parallel bases PQ and SR.

They both are equal to the altitude of the trapezoid h.

Therefore, we can represent areas of our two triangles as

S1= $\frac{1}{2}*PQ*h$

S2= $\frac{1}{2}*SR*h$

Adding them together, we get the area of the whole trapezoid:

S=S1+S2= $\frac{1}{2} (PQ+SR)h,$

which is usually represented in words as "half-sum of the bases times the altitude".

2. By dividing into different sections

Trapezoid PQRS is shown below, with PQ parallel to RS.

Figure 1 - Trapezoid PQRS with PQ parallel to RS(image is attached below.)

We are going to derive the area of a trapezoid by dividing it into different sections.

If we drop another line from Q, then we will have two altitudes namely PT and QU.

Figure 2 - Trapezoid PQRS divided into two triangles and a rectangle. (image is attached below.)

From Figure 2, it is clear that Area of PQRS = Area of PST + Area of PQUT + Area of QRU. We have learned that the area of a triangle is the product of its base and altitude divided by 2, and the area of a rectangle is the product of its length and width. Hence, we can easily compute the area of PQRS. It is clear that

=> $A_{PQRS} = (\frac{ah}{2}) + b_{1}h + \frac{ch}{2}$

Simplifying, we have

=> $A= \frac{ah+2b_{1+C} }{2}$

Factoring we have,

=> $A_{PQRS} = (a+ 2b_{1} + c)\frac{h}{2} \\= > {(a+ b_{1} + c) + b_{1} }\frac{h}{2}$

But, $a+ b_{1} + c$ is equal to $b_{2}$ , the longer base of our trapezoid.

Hence, $A_{PQRS}= (b_{1} + b_{2} )\frac{h}{2}$

We have discussed two ways by which we can derive area of a trapezoid.

Read to know more about Trapezoid

brainly.com/question/4758162?referrer=searchResults

#SPJ10

5 0

1 year ago

A rectangle has an area of (x2 − 17x + 72) square units. Since the area of a rectangle is determined using the formula, A = lw,

klasskru [66]

Answer:

length = (x − 8) units and width = (x − 9) units

Step-by-step explanation:

We need to factor

x^2 - 17x + 72

We need two numbers whose product is 72 and whose sum is -17.

The numbers are -8 and -9.

x^2 - 17x + 72 = (x - 8)(x - 9)

Answer: length = (x − 8) units and width = (x − 9) units

8 0

2 years ago

Read 2 more answers

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

Maru [420]

Parameterize $S{/tex] by[tex]\vec s(u,v)=u\,\vec\imath+v\,\vec\jmath+(8-u^2-v^2)\,\vec k$

with $0\le u\le1$ and $0\le v\le1$ .

Take the normal vector to $S$ to be

$\vec s_u\times\vec s_v=2u\,\vec\imath+2v\,\vec\jmath+\vec k$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^1\int_0^1(uv\,\vec\imath+v(8-u^2-v^2)\,\vec\jmath+u(8-u^2-v^2)\,\vec k)\cdot(2u\,\vec\imath+2v\,\vec\jmath+\vec k)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^1\int_0^1\bigg(2u^2v+(u+2v^2)(8-u^2-v^2)\bigg)\,\mathrm du\,\mathrm dv=\boxed{\frac{1553}{180}}$

6 0

3 years ago

Help me please asap

LiRa [457]

Answer:

x = 20

Step-by-step explanation:

AE is an angle bisector, which means it cuts that triangle into 2 equal halves. So this means ∠BAE and ∠EAC should be equal to each other.

Set them equal to each other and solve for x.

∠BAE = ∠EAC

x + 30 = 3x - 10

40 = 2x

x = 20

3 0

3 years ago