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3 years ago

PLEASE HELP ME ASAP!!! Thank you!

Mathematics

1 answer:

34kurt3 years ago

5 0

Original bread is 10 X 12.5 X 2. so if you cut pieces 2 X 2.5X 2 you would take 1o inches divided by 2 inches = 5 pieces (long). 12.5 inches divided by 2.5 inches wide = 5 pieces (wide). The 2 inches high cancels itself out because the bread is 2 inches high to begin with. Therefore 5 X 5 = 25 or 25 total pieces of banana bread at 2" x 2.5" x 2".

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Choose the correct product of (5x − 11)2. 25x2 − 110x 121 25x2 − 121 25x2 121 25x2 110x 121.

chubhunter [2.5K]

Product is the multiplication of numbers. The product of the (5x-11)² is (25x²-110x+121).

<h3>What is a Product?</h3>

A product is the multiplication of a number or expression to another expression.

We know in order to find the product of (5x-11)², we need to multiply each of the terms inside the bracket to each term in the second bracket. therefore, the solution of the problem can be given as,

$(5x-11)^2\\\\= (5x-11) \times (5x-11)\\\\= (5x \times 5x)-(5x \times 11)-(5x \times 11)(11 \times 11)\\\\= 25x^2 - 55x - 55x + 121\\\\= 25x -110x +121$

Hence, the product of the (5x-11)² is (25x²-110x+121).

Learn more about the Product:

brainly.com/question/6966983

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2 years ago

Find two equivalent ratios for each ratio.

Damm [24]

Answer:

Ex. 2:1 = 4:2 or 8:4

~Hope this helps!!

3 0

3 years ago

4. Are the lines parallel, perpendicular, or neither?

Lelu [443]

Answer:

Perpendicular

Step-by-step explanation:

Intersect over each other

4 0

3 years ago

Which of the following is a trinomial with a constant term?

Leto [7]

I think the answer is b

4 0

3 years ago

Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

3 0

3 years ago