55. A rhombus has side length 10 cm, and the longer diagonal measuring 10 cm.

PLEASE ANSWERRRRRRRRRR

mash [69]

Answer: OPTION B.

Step-by-step explanation:

Given the following System of equations:

$\left \{ {{3x-5y=19} \atop {x+y=1}} \right.$

You can use the Elimination Method to solve it. The steps are:

1. You can mutliply the second equation by -3.

2. Then you must add the equations.

3. Solve for the variable "y".

Then:

$\left \{ {{3x-5y=19} \atop {-3x-3y=-3}} \right.\\.....................\\-8y=16\\\\y=\frac{16}{-8}\\\\y=-2$

4. Now that you know the value of the variable "y", you must substitute it into any original equation.

5. The final step is to solve for "x" in order to find its value.

Then:

$x+(-2)=1\\\\x=1+2\\\\x=3$

Therefore, the solution is:

$(3,-2)$

8 0

4 years ago

What is 35/6 in simplest form

Len [333]

Its already in simplest form u cant simplify it hope this helps!!

3 0

4 years ago

25 POINTS AND BRAINLIEST!

Hoochie [10]

15 apple pies and lemon pies were entered.
12 apple pies and 3 lemon pies.

4 0

4 years ago

Read 2 more answers

12

allochka39001 [22]

Answer:

Let x be the rent for a movie and y is the rent for a video game

The following system of equations:

5x + 2y = 31 (1)

3x + 8y = 56 (2)

Solving the system of equations above we have:

x = (31-2y) / 5 (1) => replace x in (2) 3 ((31-2y) / 5) + 8y = 56 => y = $ 5.5

Replace y = $ 5.5 for (1) => x = ((11 - 11) / 5) = 4

So the rent for a movie is $ 5.5 and the price for renting a video games is $ 4

Step-by-step explanation:

7 0

3 years ago

For each of the following functions, determine if they are injective. Also determine if theyare surjective. Also determine if th

timurjin [86]

Answer:

Check below

Step-by-step explanation:

1. Definition for intervals

$(a,b)=\left \{ x\in\Re : a$

2. Functions

1) $\Re \rightarrow \Re \\ f(x)=x$

Let's perform graph tests.

That's an one to one, injective function. Look how any horizontal line touches that only once. Also, It's a surjective and a bijective one.

2) $\Re\geq0\rightarrow\Re , f(x)=x+1\\$

Injective, surjective and bijective.

Injective: a horizontal line crosses the graph in one point.

3) $f:\Re\geq 0\rightarrow\Re, f(x) = cos(x)$

The cosine function is not injective, bijective nor surjective.

4) $f:\Re\rightarrow\Re \:f(x)=ex$

Since e, is euler number it's a constant. It's also injective, surjective and bijective.

5) Quite unclear format

$6) \:f:\Re\rightarrow(0,\infty), f(x) =ex$

Despite the Restriction for the CoDomain, the function remains injective, surjective and therefore bijective.

$7) f:\Re\geq 0 \rightarrow \Re\geq0, f(x) =x^{4}$

Not injective nor surjective therefore not bijective too.

$8).f:\{-1,2,-3\}}\rightarrow \{1,4,9\}, f(x) =x^{2}$

$f(-1)=1, f(2)=4, f(-3)=9$

Injective (one to one), Surjective, and Bijective.

$10) f:\Re\geq 0\rightarrow [-1,1], f(x)= cos(x)\\-1=cos(x) \therefore x=\pi,3\pi,5\pi,etc.$

Surjective.

$11.f:R\geq 0[-1,1], f(x) = 0\\$

Surjective

12.f: US Citizens→Z, f(x) = the SSN of x.

General function

13.f: US Zip Codes→US States, f(x) = The state that x belongs to.

Surjective

7 0

3 years ago