Ask question

Ask question

All categories

3 years ago

12

A bakery used 25% more butter this month that last month. If the bakery used 240 kilograms of butter last month, how much did it

use this month?

2 answers:

mariarad [96]3 years ago

8 0

The answer is 300 since it’s asking the use of 25%

SCORPION-xisa [38]3 years ago

6 0

Answer:

300kg

Step-by-step explanation:

240(25/100)

240(.25) = 60

25% of 240 = 60, so add that 60 to get an extra 25% for this month

240 + 60 = 300

You might be interested in

Please help me find RS, PQ, PS, RQ, and PRQ

bulgar [2K]

The arc equals the angle so
RS= 150°
PQ= 72°
PS= 73°
RQ= 65°
PRQ= 108°

8 0

3 years ago

Read 2 more answers

Which expression is equivalent to t2 – 36?

defon

Answer:

(t + 6)(t - 6)

Step-by-step explanation:

(t - 6)(t + 6)

(t . t) + (t . (6) + ((-6) . t) + ((-6) . 6)

= t² - 6t + 6t - 36

= t² - 36

so,

t² - 36 = (t + 6)(t - 6)

6 0

3 years ago

Find the center and radius of the circle with the equation. Please show work!

mina [271]

Answer:

<h2>(1, -2)</h2>

Step-by-step explanation:

The equation of a circle:

$(x-h)^2+(y-k)^2=r^2$

(h, k) - center

r - radius

We have:

$x^2-2x+y^2+4y=4\qquad\text{add}\ 1^2\ \text{and}\ 2^2\ \text{to both sides}\\\\x^2-2x+1^2+y^2-2(-2)(y)+2^2=2(1^2)(2^2)\qquad\text{use}\ (a\pm b)^2=a^2\pm2ab+b^2\\\\(x-1)^2+(y-(-2))^2=8\\\\h=1,\ k=-2\to(1,\ -2)$

3 0

3 years ago

Whats the quotient of 15.86 divided by 0.65?

spayn [35]

Answer:

24.4

Step-by-step explanation:

15.86 /0.65 = 24.4

8 0

3 years ago

If tanA+sinA=m and tanA-sinA=n.prove that m^2-n^2=4√mn

mash [69]

Let $a=\tan A$ and $b=\sin A$ . Then

$m^2-n^2=(a+b)^2-(a-b)^2=(a^2+2ab+b^2)-(a^2-2ab+b^2)=4ab$

$\implies m^2-n^2=4\tan A\sin A$

and

$mn=(a+b)(a-b)=a^2-b^2$

$\implies4\sqrt{mn}=4\sqrt{\tan^2A-\sin^2A}$

The expression under the square root can be rewritten as

$\tan^2A-\sin^2A=\dfrac{\sin^2A}{\cos^2A}-\sin^2A=\sin^2A\left(\dfrac1{\cos^2A}-1\right)=\sin^2A(\sec^2A-1)$

Recall that

$\sin^2A+\cos^2A=1\implies\tan^2A+1=\sec^2A$

so that

$\tan^2A-\sin^2A=\sin^2A\tan^2A$

and assuming $\sin A>0$ and $\tan A>0$ , we end up with

$4\sqrt{\tan^2A-\sin^2A}=4\tan A\sin A$

so that

$m^2-n^2=4\sqrt{mn}$

as required.

5 0

3 years ago