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2 years ago

AHAISNSHKWSN HELP PLEASEEE

Mathematics

1 answer:

Lynna [10]2 years ago

8 0

Answer:

best answer is c. (2,3),(4,4),(6,5),(8,6)

Step-by-step explanation:

multiply x and y components 2*3,4*4,6*5,8*6

You might be interested in

Negative ten is no less than two times a number plus fourteen.

lawyer [7]

Is it NO less than it just less than?

5 0

4 years ago

Can someone help me out

MaRussiya [10]

Answer:

75.7

Step-by-step explanation:

Area of the sector,

17²×π×30/360

= 289π/12

= 75.7 (rounded to the nearest tenth)

Answered by GAUTHMATH

8 0

3 years ago

Q.The rectangular front windshield of the bus has the length of 1.5√3 m and the width of 1.5 m . The wipers are attached to the

galina1969 [7]

Answer:

Area covered by wiper=area of arc(A0D)+area of traingleAOB+area of arc(BOC)

We know AO and OB are wipers,whose lengths are 1.5

In ∆ABC

AB^2+BC^2=AC^2

(1.5√3)^2+(1.5)^2=(AC)^2

2.25×3+2.25=(AC)^2

2.25×4=AC^2

AC=3

Also AO=1.5

similarly using Pythagoras theorem in ∆OBC

we get

OC=1.5

this shows that point O is the bisector of diagonal AC

so,AO=OC

also from property of rectangle we know that, diagonal bisects each other equally

so,O will also be the bisector of diagonal BD

hence OB=OD=1.5

AC=BD (property of triangle)

AC/2=BD/2

OB=OC,hence OB=OC=OD=OA=1.5=AD=BC

Hence OAD will be an equilateral triangle

angle OAD=60

similarly angleOBC=60

area of sector=angle of that sector ×area of circle/360

area of arc(AOD)+arc(BOC)=60/360 ×3.14 ×1.5×1.5 +60/360 ×3.14×1.5×1.5

=2×(60/360)×1.5×1.5

area of triangleAOB=area of rectangle/4 (diagonal divides the rectangle in 4 equal parts)

area of triangle=1.5×1.5√3/4

=>1.5×2.6/4=3.90/4

=0.975 or 0.98(approx)

Area covered by wiper=.98+2×60/360 ×3.14×1.5×1.5

=.98+0.75×3.14=2.35+0.97

=2.35+.98

Percentage of area of windshield covered by wiper=3.3×100/3.9=84.61

hence percentage of area of windshield covered by wiper is 85percent(approx)

plz mark brainlest

5 0

3 years ago

A right triangle has side lengths 7, 24, and 25 as shown below.

Hoochie [10]

Answer: Tan B = 0.2917 (16.26 degrees)

Sin B = 0.2800 (16.26 degrees)

Cos B = 0.9600 (16.26 degrees)

Step-by-step explanation: Please refer to the picture attached for details.

The right angled triangle LNB has been drawn with the sides labeled as 7, 24 and 25 as indicated. With angle B as the reference angle, we can now identify each side as follows;

Side 7 (opposite, facing the reference angle)

Side 24 (adjacent, between the reference angle and the right angle)

Side 25 (hypotenuse, facing the right angle).

Hence, using the trigonometric ratios,

Tan B = opposite/adjacent

Tan B = 7/24

Tan B = 0.2917 (16.26 degrees)

Sin B = opposite/hypotenuse

Sin B = 7/25

Sin B = 0.2800 (16.26 degrees)

Cos B = adjacent/hypotenuse

Cos B = 24/25

Cos B = 0.9600 (16.26 degrees)

3 0

4 years ago

(GEOMETRY PLS HELP SERIOUSLY) A treasure map says that a treasure is buried so that it partitions the distance between a rock an

pogonyaev

You have coordinates of tree T(16,21) and of rock R(3,2). Let S be a point that denotes treasure. Since treasure is buried so that it partitions the distance between a rock and a tree in a 5:9 ratio, you can state that

$\dfrac{\overrightarrow{RS}}{\overrightarrow{ST}}=\dfrac{5}{9}.$

Find the coordinates of these vectors:

$\overrightarrow{RS}=(x_S-3,y_S-2), \\
\overrightarrow{ST}=(16-x_S,21-y_S).$

Then

$\dfrac{x_S-3}{16-x_S} =\dfrac{5}{9}$ and $\dfrac{y_S-2}{21-y_S} =\dfrac{5}{9}.$

Solve these two equations:

$9(x_S-3)=5(16-x_S),\\ 9x_S-27=80-5x_S,\\ 9x_S+5x_S=80+27,\\14x_S=107,\\ \\x_S=\dfrac{107}{14}\approx 7.6.$

$9(y_S-2)=5(21-y_S),\\9y_S-18=105-5y_S,\\9y_S+5y_S=105+18,\\14y_S=123,\\ \\y_S=\dfrac{123}{14}\approx 8.8.$

Answer: the treasure are placed at point (7.6,8.8), so the correct choice is B.

5 0

4 years ago

Read 2 more answers