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3 years ago

(4x+5)(4x−5) I'm stuck can someone please help me

Mathematics

2 answers:

Marizza181 [45]3 years ago

5 0

Answer:6small2−25

Step-by-step explanation:

Vesnalui [34]3 years ago

3 0

Answer:

16x² - 25

Step-by-step explanation:

(4x+5)(4x−5)

16x² - 20x + 20x - 25

16x² - 25

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The area of a isosceles trapezoid height of 8 base of 10 base of 15

ikadub [295]

Answer:

A = 100 $un^{2}$

Step-by-step explanation:

You have to use the area of a trapezoid formula:

A = $\frac{h}{2}$ (a + b)

Now, substitute the values of the height and the bases:

A = $\frac{8}{2}$ (10 + 15)

Finally, simplify the equation to solve for the area:

A = 4(25)

A = 100 $un^{2}$

6 0

3 years ago

The United States five-dollar bill are 155.956 mm long and 66.294 wide. A stack of these bills fits inside a 156 x 66.3 x 66.3 m

maxonik [38]

Answer:

A) $5 × 25 = $75

B) 37.69%

Step-by-step explanation:

The dimension of five dollar bill:

Length = 155.956mm

Width = 66.294

Volume of box being occupied by 5 dollar bill = 258,473.68mm^3

Dimension of box :

156 x 66.3 x 66.3 mm

If volume of the 5—dollar bill is 258,473.68mm^3

Then, the Thickness of the five dollar stack equals :

Volume = Length × width × height

258,473.68mm^3 = 155.956mm × 66.294 × height

258,473.68 = 10338.947064 × h

h = (258473.68/10338.947064)

h = 25.000000003288 = 25

Therefore, the amount of money in the box is:

$5 × 25 = $75

Percentage of box volume taken up by $5 bill

Volume of the box:

156 x 66.3 x 66.3 mm = 685727.6399mm^3

Volume of box being occupied by 5 dollar bill = 258,473.68mm^3

(258,473.68 / 685727.6399) × 100

=37.69%

4 0

3 years ago

Solve system of equations for x + y: -4x + 9y = -11 & -3x + 7y =-9

Aloiza [94]

Answer:

x = -3 and y = -4

x + y = -3-4 = -7

Step-by-step explanation:

-4x + 9y = -11 ------------------------------------------------------------------------(1)

-3x + 7y = -9--------------------------------------------------------------------------(2)

multiply through equation (1) by 3 and multiply through equation(2) by 4

-12x + 27y = -33 -----------------------------------------------------------------------(3)

-12x + 28y = -36 ----------------------------------------------------------------------(4)

subtract equation (1) from equation (2)

y = -3

substitute y = -3 into equation (2)

-3x + 7(-3) = -9

-3x -21 = -9

add 21 to both-side of the equation

-3x -21 + 21 = -9 + 21

-3x = 12

divide both-side of the equation by -3

-3x/-3 = 12/-3

x = -4

x = -3 and y = -4

8 0

3 years ago

The price of a home is $120,000. The bank requires a 10% down payment and two points at the time of closing. The cost of the hom

rusak2 [61]

Answer:

The amount of the loan is $120,000 -10% + 2% = $110,400, assuming the "points" are added to the financed amount.

I find that the monthly payment will be

$848.89. There will be 360 such payments over 30 years, for a total payment amount of $30The amount of the loan is $120,000 -10% + 2% = $110,400, assuming the "points" are added to the financed amount.

R = Pi/[1 - (1+i)^(-n)] where

P = 110,400

R = the monthly payment

i = 8.5/[100(12)] = .007083...

n = 30(12) = 360 yielding

R = $848.88 per month.

Therefore, the total interest paid is 360(848.88) - 100,400.

5,600. Subtract the original principle (110,400) from that to get the interest paid.

hopes this helps :]

4 0

2 years ago

Choose the best coordinate system to find the volume of the portion of the solid sphere rho <_4 that lies between the cones φ

MrRissso [65]

Answer:

So, the volume is:

$\boxed{V=\frac{128\sqrt{2}\pi}{3}}$

Step-by-step explanation:

We get the limits of integration:

$R=\left\lbrace(\rho, \varphi, \theta):\, 0\leq \rho \leq 4,\, \frac{\pi}{4}\leq \varphi\leq \frac{3\pi}{4},\, 0\leq \theta \leq 2\pi\right\rbrace$

We use the spherical coordinates and we calculate a triple integral:

$V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\int_0^4 \rho^2 \sin \varphi \, d\rho\, d\varphi\, d\theta\\\\V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sin \varphi \left[\frac{\rho^3}{3}\right]_0^4\, d\varphi\, d\theta\\\\V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sin \varphi \cdot \frac{64}{3} \, d\varphi\, d\theta\\\\V=\frac{64}{3} \int_0^{2\pi} [-\cos \varphi]_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \, d\theta\\\\V=\frac{64}{3} \int_0^{2\pi} \sqrt{2} \, d\theta\\\\$

we get:

$V=\frac{64}{3} \int_0^{2\pi} \sqrt{2} \, d\theta\\\\V=\frac{64\sqrt{2}}{3}\cdot[\theta]_0^{2\pi}\\\\V=\frac{128\sqrt{2}\pi}{3}$

So, the volume is:

$\boxed{V=\frac{128\sqrt{2}\pi}{3}}$

4 0

3 years ago

(4x+5)(4x−5) I'm stuck can someone please help me ​

(4x+5)(4x−5) I'm stuck can someone please help me