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3 years ago

Can y’all please help

Mathematics

1 answer:

xxMikexx [17]3 years ago

4 0

Given: f(x) = 2x^2 +5sqrt(x-2)

Evaluate this for x=6:

f(6) = 2(6)^2 + 5sqrt(6-2) = 72 + 5(2) = 72 + 10 = 82 (answer)

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I need help asap. Evaluate Expressions

Anit [1.1K]

Answer:

p-r/6=-6-(-6)/6=-6+1=-5

2:

p-(m-n)= -1-(-4-4)=-1-(-8)=-1+8=7

(z+y)/2)3=(-5-4)/3=-9/3=-3

4:

m/6-n=6/6-6=1-6=-5

k³-h=3³-(-2)=27+2=29

p-(p-(m-3))=5-(5-(4-3))=5-(5-1)=5-4=1

(k)(k-j)+3=(5)(5-5)+3=5*0+3=3

p²/6-q=6²/6-4=6-4=2

9;

zx+3³=6*4+3³=24+27=51.

10:

y+z-(y-x)=5+4-(5-2)=9-3=6

Some basic rules:

+. +. +=+

+. * +=+

+. ÷ +=+

+. - +=+

and

+. +. -=add and put sigh of greater one.

-. * +=-sigh

-÷-=+sigh

- - -=add and put - sigh

7 0

3 years ago

Read 2 more answers

Please help plzzzzzzzzzzzzzzzzzzzz

zepelin [54]

Answer: C,D are true

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

A plane is descending into the airport. After 5 minutes it is at a height of 6500 feet. After 7 minutes it is at a height of 590

Jet001 [13]

Let's assume

height of plane in feet =h

time in minutes =t

we are given

A plane is descending into the airport. After 5 minutes it is at a height of 6500 feet

so, we get one point as (5,6500)

After 7 minutes it is at a height of 5900 feet

so, we get another point as (7,5900)

we can use point slope form of line

$y-y_1=m(x-x_1)$

points as

(5,6500)

x1=5, y1=6500

(7,5900)

x2=7 , y2=5900

Calculation of slope(m):

$m=\frac{y_2-y_1}{x_2-x_1}$

now, we can plug values

$m=\frac{5900-6500}{7-5}$

$m=-300$

Equation of line:

we can use formula

$y-y_1=m(x-x_1)$

we can plug values

$h-6500=-300(t-5)$

$h=-300t+8000$

Time of landing:

we can set h=0

and then we can solve for t

$h=-300t+8000=0$

$t=\frac{80}{3}min$ ..............Answer

5 0

3 years ago

180kg, 250kg, 200kg, 209kg, 195kg, 205kg, 190kg, 188kg, 192kg. What is the IQR of the lion's weights.

velikii [3]

Answer:

Step-by-step explanation:

Given the data:

180kg, 250kg, 200kg, 209kg, 195kg, 205kg, 190kg, 188kg, 192kg

The interquartile range (IQR) = Q3 - Q1

Reordering the data:

180, 188, 190, 192, 195, 200, 205, 209, 250

Q3 = 3/4(n+1)th term

Q3 = 3/4(10) = 7.5 th term = (205+209)/2 = 207

Q1 = 1/4(n+1)th term

Q1 = 1/4(10) = 2.5th term = (188+190)/2 = 189

Q3 - Q1 = 207 - 189 = 18

3 0

3 years ago

Help me to do 10(b) answer

antiseptic1488 [7]

Answer:

Step-by-step explanation:

Big circle:

R = radius = diameter ÷2 = 42 ÷ 2 = 21 cm

Area of big circle= πR²

$=\dfrac{22}{7}*21*21=22*3*21\\\\\\= 1386 \ cm^{2}$

Small circle:

Diameter of small circle = radius of big circle = 21 cm

r = 21/2 = 10.5 cm

Area of small circle = πr²

$=\dfrac{22}{7}*10.5*10.5 = 22 * 1.5*10.5\\\\\\= 346 .5 \ cm^{2}$

Area of shaded region = area of big circle - area of small circle

= 1386 - 346.5

= 1039.5 cm²

3 0

2 years ago