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4 years ago

Can someone please help me with this question

Mathematics

1 answer:

Anton [14]4 years ago

5 0

It seems the first line has to be

1. AB=CD and AD=CB 1. given

2. AC=AC 2. Reflexivity (things are equal to themselves)

3 $\triangle$ ACB $\cong$ $\triangle$ CAD 3. SSS

4. $\angle$ 1 = $\angle$ 4. Corresponding parts of congruent triangles

5 DC || AB 5. Congruent alternate traversal angles imply parallel lines

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Jamal ran 8 kilometers, sprinted 500 meters, and climbed bleachers for a distance of 1.5 kilometers during track practice. How m

Karo-lina-s [1.5K]

In total, he ran 10,000 meters. The way you can find this is by finding how many meters are in a kilometer (which is 1,000), after that you need to convert everything to meters, so running 8km is now running 8,000 meters, sprinting 500 meters doesn't change, and climbing bleachers for 1.5km is now 1,500 meters.
If you add all of these numbers up you will get your answer of 10,000 meters

4 0

4 years ago

PLEASE HELP ASAP, I WILL GIVE BRAINLESSLY ANSWER SHOW WORK PLEASE

zhannawk [14.2K]

Answer:

C. $N(t)=150\cdot 3^t$

Step-by-step explanation:

You are given the exponential function $n(t)=ab^t.$

From the table, $N(t)=150$ at $t=0,$ thus

$N(0)=a\cdot b^0\\ \\150=a\cdot 1\ [\text{ because }b^0=1]$

Also $N(t)=450$ at $t=1,$ thus

$N(1)=a\cdot b^1=a\cdot b.$

Since $a=150,$ substitute it into the second equation

$450=150\cdot b\\ \\b=\dfrac{450}{150}\\ \\b=3$

and the expression for the exponential function is

$N(t)=150\cdot 3^t$

5 0

4 years ago

2x -6y = 5 x+y=2 Solve the system of equations. (17, -15) (17/8, -18) no solution

yan [13]

Answer:

(17/8, -1/8)

Step-by-step explanation:

$\left \{ {{2x-6y=5} \atop {x+y=2}} \right.$

the bottom equation is equivalent to x = 2-y

so I switch the value of x in the top equation to x = 2-y

and it becomes 2*(2-y) - 6y = 5

equal to 4 - 2y - 6y = 5

-8y = 1

y = -1/8

then you substitute this value in the second equation

x + y = 2 becomes x + (-1/8) = 2

x = 2 + 1/8 = 16/8 + 1/8 = 17/8

6 0

4 years ago

10 POINTS + BRAINLIEST ANSWER!!

svp [43]

C. tan P is the answer to your question

8 0

4 years ago

Read 2 more answers

Show that an integer is divisible by 11 if and only if the alternating sum (add first digit, subtract the second, add the third,

Bess [88]

Answer and Step-by-step explanation:

Suppose that we have a number y which is a positive integer and that:

y = $x_n...x_5x_4x_3x_2x_1x_0$

Where;

$x_{0}$ = digit at 10⁰ => one's place (units place)

$x_1$ = digit at 10¹ => 10's place (tens place)

$x_{2}$ = digit at 10² => 100's place (hundreds place)

$x_{3}$ = digit at 10³ => 1000's place (thousands place)

$x_{n}$ = digit at 10ⁿ place

Then;

y = $x_{0}$ * 10⁰ + $x_1$ * 10¹ + $x_{2}$ * 10² + $x_{3}$ * 10³ + $x_{4}$ * 10⁴ + $x_5$ * 10⁵ + . . . + $x_{n}$ * 10ⁿ

Since 10⁰ = 1, let's rewrite y as follows;

y = $x_{0}$ + $x_1$ * 10¹ + $x_{2}$ * 10² + $x_{3}$ * 10³ + $x_{4}$ * 10⁴ + $x_5$ * 10⁵ + . . . + $x_{n}$ * 10ⁿ

Now, to test if y is divisible by 11, replace 10 in the equation above by -1. Since 10 divided by 11 gives -1 (mod 11) [mod means modulus]

y = $x_{0}$ + $x_1$ * (-1)¹ + $x_{2}$ * (-1)² + $x_{3}$ * (-1)³ + $x_{4}$ * (-1)⁴ + $x_5$ * (-1)⁵ + . . . + $x_{n}$ * (-1)ⁿ

=> y = $x_{0}$ - $x_1$ + $x_{2}$ - $x_{3}$ + $x_{4}$ - $x_5$ + . . . + $x_{n}$ (-1)ⁿ (mod 11)

Therefore, it can be seen that, y is divisible by 11 if and only if alternating sum of its digits $x_{0}$ - $x_1$ + $x_{2}$ - $x_{3}$ + $x_{4}$ - $x_5$ + . . . + $x_{n}$ (-1)ⁿ is divisible by 11

Let's take an example

Check if the following is divisible by 11.

i. 1859

Solution

1859 is divisible by 11 if and only if the alternating sum of its digit is divisible by 11. i.e if (1 - 8 + 5 - 9) is divisible by 11.

1 - 8 + 5 - 9 = -11.

Since -11 is divisible by 11 so is 1859

ii. 31415

Solution

31415 is divisible by 11 if and only if the alternating sum of its digit is divisible by 11. i.e if (3 - 1 + 4 - 1 + 5) is divisible by 11.

3 - 1 + 4 - 1 + 5 = 10.

Since 10 is not divisible by 11 so is 31415 not divisible.

3 0

4 years ago