All categories

3 years ago

Can you please tell me the answer I don't know

Mathematics

1 answer:

9966 [12]3 years ago

8 0

Answer:

Twenty-eight billion five hundred forty-two million rupees

28,542,000,000 rupess

or, 28.542 × 10⁹ rupees

You might be interested in

Andrea constructed a triangle. Angle 1 and 3 are the same size and angle 2 has a measurement of 70 degrees. What is the measurem

Verdich [7]

Answer:

Step-by-step explanation:

By triangle sum theorem,

Sum of all angles of a triangle is 180°.

m∠1 + m∠2 + m∠3 = 180°

(m∠1 + m∠3) + m∠2 = 180°

2(m∠1) + 70° = 180° {Given → m∠1 = m∠3]

2(m∠1) = 110°

m∠1 = 55°

Therefore, m∠1 = m∠3 = 55°

7 0

3 years ago

PLEASE HURRY!!! ONLY HAVE A FEW MINS

aleksley [76]

Answer:

it is 13x=2

Step-by-step explanation:

i ran the math through my multipurpose calculator

7 0

3 years ago

Read 2 more answers

I need to know how to write this expression forty-three less than twice a number, n

Sindrei [870]

N= a number
2n= twice a number

If it's 43 less, then it would be 2n-43.

2n-43 is the answer.

Hope this helps! :)

5 0

3 years ago

Read 2 more answers

Tell whether the lines for each pair of questions are parallel or perpendicular or neither y=5/3x+3. 20x+12y=12

disa [49]

neither

• Parallel lines have equal slopes

• The product of perpendicular slopes = - 1

the equation of a line in slope-intercept form is

y = mx + c ( m is the slope and c the y-intercept )

y = $\frac{5}{3}$ x + 3 is in this form

with slope m = $\frac{5}{3}$

rearrange 20x + 12y = 12 into this form

subtract 20x from both sides

12y = - 20x + 12 ( divide all terms by 12 )

y = - $\frac{5}{3}$ + 1 ← in slope-intercept form

with slope m = - $\frac{5}{3}$

Neither of the conditions for parallel/ perpendicular slopes are met

Hence the lines are neither parallel/ perpendicular

8 0

3 years ago

Look at this cylinder:

Sedbober [7]

Height=h=8cm
Radius=r=4cm

We know

$\boxed{\sf \star TSA_{(Cylinder)}=2\pi r(h+r)}$

$\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=2\times \dfrac{22}{7}\times 4(8+4)$

$\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=\dfrac{176}{7}(12)$

$\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=\dfrac{2112}{7}$

$\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=301.7cm^2$

Now

New Radius=2(4)=8cm
New Height=2(8)=16cm

$\\ \sf\longmapsto TSA_{(New\:Cylinder)}=2\times \dfrac{22}{7}\times 8(16+8)$

$\\ \sf\longmapsto TSA_{(New\:Cylinder)}=\dfrac{352}{7}(24)$

$\\ \sf\longmapsto TSA_{(New\:Cylinder)}=\dfrac{8448}{7}$

$\\ \sf\longmapsto TSA_{(New\:Cylinder)}=1204.7cm^2$

$\\ \sf\longmapsto \dfrac{TSA_{(New\:Cylinder)}}{TSA_{(Old\:Cylinder)}}=\dfrac{1204.7}{301.7}$

$\\ \sf\longmapsto \dfrac{TSA_{(New\:Cylinder)}}{TSA_{(Old\:Cylinder)}}=\dfrac{4}{1}$

$\\ \sf\longmapsto\underline{\boxed{\bf{ {TSA_{(New\:Cylinder)}}:{TSA_{(Old\:Cylinder)}}=4:1}}}$

Hence our correct option is Option C

6 0

2 years ago

Can you please tell me the answer I don't know ​

Can you please tell me the answer I don't know