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3 years ago

6

Use the digits 3,7,1 and 5 to write a number that rounds to 1,700

2 answers:

Hitman42 [59]3 years ago

5 0

To how many sig figs or decimal places? Could use 1735 if rounded to 2 significant figures (2sf)

Ray Of Light [21]3 years ago

3 0

1735 is the answer to your problem because 1 is in the 1000 place, 7 is in the 100 place, 3 is in the 10 place, and 5 is in the 1 place which would round to 1700

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Two poles, AB and ED, are fixed to the ground with the help of ropes AC and EC, as shown:

kipiarov [429]

Answer:

15.59 feet

Step-by-step explanation:

We need to find BC and CD

First find BC

We can use the Pythagorean theorem

AB^2 + BC^2 = AC^2

11^2 + BC^2 = 13^2

121 + BC^2 =169

Subtract 121 from each side

121-121+BC^2 = 169-121

BC^2 = 48

Take the square of each side

sqrt(BC^2 ) = sqrt( 48)

BC = 6.92820323

then we need to find CD

We can use the Pythagorean theorem

CD^2 + DE^2 = CE^2

CD^2 + 5^2 = 10^2

CD^2 + 25 =100

Subtract 25 from each side

CD^2 + 25-25 =100-25

CD^2 = 75

Take the square of each side

sqrt(CD^2 ) = sqrt( 75)

BC = 8.660254038

BD = BC +CD

= 6.92820323 + 8.660254038

15.58845727

7 0

3 years ago

a rectangle is transformed according to the rule r0, 90º. the image of the rectangle has vertices located at r'(–4, 4), s'(–4, 1

AleksandrR [38]

Graph images undergo transformations, one of the types of transformations are rotations.
rotation is done when each point of the image rotates 90° around a point in the counterclockwise manner.
when rotation is done, (x,y) -> (-y,x)
therefore when taking points from image to original then the reverse is done ;
(-y,x) --> (x,y)
the coordinates of image point q' is (-3,4)
before rotation then the coordinates are (4,3)
answer is (4,3)

4 0

3 years ago

Read 2 more answers

A man walked 6 km per hour how far does he work between 9:30 a.m. and 12 p.m.

kolbaska11 [484]

Answer:

15 km

Step-by-step explanation:

9:30 to 12:00 = 2.5 hours

2.5 x 6 = 15

8 0

2 years ago

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igor_vitrenko [27]

Answer:

30

Step-by-step explanation:

3 0

3 years ago

Can you answer these please many thanks

givi [52]

Given: (a) $2(x+3)=2x+6$ and (b) $4(y-3)=4y-12$ .

To find: The expanded form of the given expressions.

(c) $4(m+n)=4m+4n$ (using $a(b+c)=ab+ac$ )

(d) $3(5-q)=15-3q$ (using $a(b-c)=ab-ac$ )

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(f) $3(2x-5)=6x-15$ (using $a(b-c)=ab-ac$ )

(g) $7(4b-1)=28b-7$ (using $a(b-c)=ab-ac$ )

(h) $3(2x+y-5)=3((2x+y)-5)$

⇒ $3(2x+y-5)=3(2x+y)-15$ (using $a(b-c)=ab-ac$ )

⇒ $3(2x+y-5)=6x+3y-15$ (using $a(b+c)=ab+ac$ )

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⇒ $2(6a-4b+3)=2(6a-4b)+6$ (using $a(b+c)=ab+ac$ )

⇒ $2(6a-4b+3)=12a-8b+6$ (using $a(b-c)=ab-ac$ )

(j) $6(m+n+p)=6((m+n)+p)$

⇒ $6(m+n+p)=6(m+n)+6p$ (using $a(b+c)=ab+ac$ )

⇒ $6(m+n+p)=6m+6n+6p$ (using $a(b+c)=ab+ac$ )

(k) $y(y+2)=y^2+2y$ (using $a(b+c)=ab+ac$ )

(l) $g(g-3)=g^2-3g$ (using $a(b-c)=ab-ac$ )

(m) $n(4-n)=4n-n^2$ (using $a(b-c)=ab-ac$ )

(n) $a(b+c)=ab+ac$ (using $a(b+c)=ab+ac$ )

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(q) $4y(x-3)=4xy-12y$ (using $a(b-c)=ab-ac$ )

(r) $5a(2b-5)=10ab-25a$ (using $a(b-c)=ab-ac$ )

(s) $4a(3b+2c)=12ab+8ac$ (using $a(b+c)=ab+ac$ )

(t) $5p(4p-5q)=20p^2-25pq$ (using $a(b-c)=ab-ac$ )

6 0

3 years ago