All categories

3 years ago

NEED HELP ASAP ON GEOMETRY!! ONLY 3 QUESTIONS ATTACHED WITH PHOTOS. THANK YOU

Mathematics

1 answer:

ankoles [38]3 years ago

4 0

3. Look at the picture.

Use Pythagorean theorem:

$x^2=3.6^2+5.5^2\\\\x^2=12.96+30.25\\\\x^2=43.21\to x=\sqrt{43.21}\\\\\boxed{x\approx6.6}$

4.
Use Pythagorean theorem:

$3^2+\left(\dfrac{x}{2}\right)^2=6.5^2\\\\9+\dfrac{x^2}{4}=42.25\ \ \ |-9\\\\\dfrac{x^2}{4}=33.25\ \ \ \ |\cdot4\\\\x^2=133\to x=\sqrt{133}\\\\\boxed{x\approx11.5}$

5.
$\widehat{AB}\ \cong\ \widehat{DE}$

You might be interested in

The school that Scott goes to is selling tickets to a fall musical. On the first day of ticket sales the school sold 12 adult ti

harina [27]

Answer:

The price of one adult ticket = $13

The price of one student ticket = $4

Step-by-step explanation:

Let the price of 1 adult ticket = x

Let the price of 1 student ticket = y

On the first day of ticket sales the school sold 12 adult tickets and 10 student tickets for a total of $196.

12 × x + 10 × y = $196

12x + 10y = 196....... Equation 1

The school took in $59 on the second day by selling 3 adult tickets and 5 student tickets

3 × x + 5 × y = $59

3x + 5y = 59.......... Equation 2

Using Elimination method

We eliminate y, by Multiplying equation 1 by 5 and Equation 2 by 10

12x + 10y = 196....... Equation 1 × 5

3x + 5y = 59.......... Equation 2 × 10

60x + 50y = 980....... Equation 3

30x + 50y = 590......... Equation 4

Subtract Equation 4 from Equation 3

30x = 390

x = 390/30

x = $13

Therefore, the price of one adult ticket = $13

Remember: x = $13

3x + 5y = 59.......... Equation 2

Substitute

3(13) + 5y = 59

39 + 5y = 59

5y = 59 - 39

5y = 20

y = 20/5

y = $4

Therefore, the price of one student ticket = $4

8 0

2 years ago

Mathew walks 5/8 mile to the bus stop each morning.How far will he walk in 5 days

LenaWriter [7]

Answer:

3 1/8 miles I think

Step-by-step explanation:

5/8 x 5/1 = 25/8

Divide 25 by 8 and you get 3 1/8

Add label of miles

7 0

2 years ago

Let f(x) = -3x + 2 and g(x) - 4x + 6. Find (f•g)(-5).

mezya [45]

Answer:

(f•g)(-5) = -238

Step-by-step explanation:

We have that:

f(x) = -3x + 2

g(x) = 4x + 6

Find (f•g)(-5).

First we find (f.g)(x)

f(x)*g(x) = (-3x + 2)*(4x + 6) = -12x² -10x + 12

When x = -5

-12(-5)² - 10(-5) + 12 = -238

(f•g)(-5) = -238

6 0

3 years ago

jacob wants to buy packets of fruit juice. on tuesday, he buys 8 packets of orange and 6 packets of Apple the packets make a tot

ioda

Answer:

Each packet of orange can make 2 cups of juice and each packets of Apple can make 3 cups of juice.

Step-by-step explanation:

Let number of cups of juice in each packet of orange be 'x'.

Let number of cups of juice in each pack of apples be 'y'.

Given:

On Tuesday;

Number of packets of oranges = 8

Number of packets of apples = 6 packets.

Amount of fruit juice = 34 cups

So we can say that;

Amount of fruit juice is equal to sum of Number of packets of oranges multiplied by number of cups of juice in each packet of orange and Number of packets of apples multiplied by number of cups of juice in each pack of apples.

framing in equation form we get;

$8x+6y=34 \ \ \ \ equation \ 1$

On Wednesday;

Number of packets of oranges = 14

Number of packets of apples = 6

Amount of fruit juice = 46 cups

So we can say that;

framing in equation form we get;

$14x+6y=46 \ \ \ \ equation \ 2$

Now Subtracting equation 1 from equation we get;

$14x+6y-(8x+6y)=46-34\\\\14x+6y-8x-6y=12\\\\6x=12$

Now dividing both side by 6 we get;

$\frac{6x}{6}=\frac{12}{6}\\\\x=2\ cups$

Now we will substitute the value of 'x' in equation 1 we get;

$8x+6y=34\\\\8\times2+6y=34\\\\16+6y=34$

Subtracting both side by 16 we get;

$16+6y-16=34-16\\\\6y= 18$

Now dividing both side by 6 we get;

$\frac{6y}{6}=\frac{18}{6}\\\\y=3$

Hence Each packet of orange can make 2 cups of juice and each packets of Apple can make 3 cups of juice.

5 0

3 years ago

Expansion Numerically Impractical. Show that the computation of an nth-order determinant by expansion involves multiplications,

posledela

Answer:

number of multiplies is n!
n=10, 3.6 ms
n=15, 21.8 min
n=20, 77.09 yr
n=25, 4.9×10^8 yr

Step-by-step explanation:

Expansion of a 2×2 determinant requires 2 multiplications. Expansion of an n×n determinant multiplies each of the n elements of a row or column by its (n-1)×(n-1) cofactor determinant. Then the number of multiplies is ...

mpy[n] = n·mp[n-1]

mpy[2] = 2

So, ...

mpy[n] = n! . . . n ≥ 2

If each multiplication takes 1 nanosecond, then a 10×10 matrix requires ...

10! × 10^-9 s ≈ 0.0036288 s ≈ 0.004 s . . . for 10×10

Then the larger matrices take ...

n=15, 15! × 10^-9 ≈ 1307.67 s ≈ 21.8 min

n=20, 20! × 10^-9 ≈ 2.4329×10^9 s ≈ 77.09 years

n=25, 25! × 10^-9 ≈ 1.55112×10^16 s ≈ 4.915×10^8 years

_____

For the shorter time periods (less than 100 years), we use 365.25 days per year.

For the longer time periods (more than 400 years), we use 365.2425 days per year.

8 0

3 years ago