Ask question

Ask question

All categories

3 years ago

13

a number consists of two digits. One of the digit is greater than the other by 5 when the digits are reversed the number become

3 divided by 8 of the original number. find the number

1 answer:

Rom4ik [11]3 years ago

6 0

Answer:

the number is 27

Step-by-step explanation:

7 - 2 = 5

72 ÷ 3 = 24

24 ÷ 8 = 3

You might be interested in

5x-8=37 find the value of x

Andrej [43]

Because <span>−8</span><span> does not contain the </span>variable<span> to solve for move it to the other side of the </span>equation<span> by adding </span>8<span> to both sides.
</span><span>5x=8+37

</span>Add 8<span> and </span>37<span> to get </span><span>45.
</span><span>5x=45

</span>Divide<span> each </span>term<span> by 5.
</span>x=<span>9
</span>

x is equal to 9

3 0

3 years ago

Pls help will give brainliest

cupoosta [38]

Answer:

the book club has a greater mean than the travel club

Step-by-step explanation:

3 0

2 years ago

SOMEONE HELP ME IM FREAKING OUT I LITERALLY CANT WITH THIS QUESTION IM PRAYING PLEASE HELP ME IM SO SERIOUS IM GONNA END IT PLS

antiseptic1488 [7]

Answer:

$\sf -11+7\sqrt{2}$

Step-by-step explanation:

Given expression:

$\sf \dfrac{3-\sqrt{32}}{1+\sqrt{2} }$

Rewrite 32 as 16 · 2:

$\sf \implies \dfrac{3-\sqrt{16 \cdot 2}}{1+\sqrt{2} }$

Apply radical rule $\sf \sqrt{a \cdot b}=\sqrt{a}\sqrt{b}$

$\sf \implies \dfrac{3-\sqrt{16}\sqrt{2}}{1+\sqrt{2} }$

As $\sf \sqrt{16}=4$ :

$\sf \implies \dfrac{3-4\sqrt{2}}{1+\sqrt{2} }$

Multiply by the conjugate:

$\sf \implies \dfrac{3-4\sqrt{2}}{1+\sqrt{2} } \times \dfrac{1-\sqrt{2} }{1-\sqrt{2} }$

$\sf \implies \dfrac{(3-4\sqrt{2})(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})}$

$\sf \implies \dfrac{3-3\sqrt{2}-4\sqrt{2}+4\sqrt{2}\sqrt{2}}{1-\sqrt{2}+\sqrt{2}-\sqrt{2}\sqrt{2}}$

As $\sf \sqrt{2}\sqrt{2}=\sqrt{4}=2$ :

$\sf \implies \dfrac{3-3\sqrt{2}-4\sqrt{2}+4 \cdot 2}{1-\sqrt{2}+\sqrt{2}-2}$

$\sf \implies \dfrac{3-7\sqrt{2}+8}{1-2}$

$\sf \implies \dfrac{11-7\sqrt{2}}{-1}$

$\sf \implies -11+7\sqrt{2}$

7 0

2 years ago

L need the answer for this question asap!!!

DochEvi [55]

<h2>Answer:</h2><h3>A. Domain </h3>

The domain of a function is the x-values that the graph applies to. This means that the domain is whatever x-values the graph crosses. All vertical parabolas (like the one pictured) have a domain of all reals. This is because any x-value could be plugged into the function and provide a y-value. while it may not seem like it, that graph will cover every single x-value in existence.

Domain = All reals

<h3>B. Range</h3>

The range is similar to the domain but is for y-values. So, the range is whatever y-values the graph applies to and crosses. As you can see from the graph, there are no y-values above 1. This means the range is y≤1.

Range = y ≤ 1

<h3>C. Increasing Interval</h3>

A graph is increasing when the y-values are increasing. So, on the parent function of a parabola, the graph increases to the right and decreases to the left. However, this graph is inverted and shifted to the left, so the interval will also be flipped and shifted. In this case, the graph increases from -∞ to 2.

Increasing Interval = [-∞, 2]

<h3>D. Decreasing Interval</h3>

The decreasing interval is very similar to the increasing interval. This interval applies when the y-values are decreasing as the x-values increase. For a parabola, the increasing and decreasing intervals always meet at the x-value of the vertex, which is 2 on this graph. The y-values decrease during the interval 2 to ∞.

Decreasing Interval = [2, ∞]

<h3>E. Opening</h3>

The direction of a parabola is decided by the sign (+ or -) of the leading coefficient. Positive coefficients open up and negative opens down. As you can see from the graph, the sides of the parabola point downwards. This means that the leading coefficient must be negative.

Opening = Down

<h3>F. Min and Max</h3>

A parabola will always only have a min or a max, never both. If a graph opens up it has a min because there is one y-value which is the minimum possible y-value. Graphs that open downwards have a maximum because there is one y-value that is the largest possible. So, this graph has a maximum of 1 because that is the largest possible y-value.

Max = 1

4 0

2 years ago

What’s is part A?? .....

bixtya [17]

Answer:

A. 3

B. 4

C. 1

D. 2

Step-by-step explanation:

Consider all equations:

A. $y=x^2 -6x+8$

This is the equation of parabola with vertex at point

$x_v=\dfrac{-b}{2a}=\dfrac{-(-6)}{2\cdot 1}=3\\ \\y_v=3^2-6\cdot 3+8=9-18+8=-1$

The y-intercept is at point

$x=0\\ \\y=0^2-6\cdot 0+8=8$

Since

$x^2 -6x+8=x^2-4x-2x+8=x(x-4)-2(x-4)=(x-2)(x-4)$ ,

the x-intercepts are at points (2,0) and (4,0)

The leading coefficient is 1 > 0, then the parabola opens upwards.

Hence, the graph of this parabola is 3.

B. $y=(x-6)(x+8)$

This parabola has two x-intercepts at points (6,0) and (-8,0).

The leading coefficient is 1 > 0, then the parabola opens upwards.

The only possible choice is parabola 4.

C. $y=(x-6)^2+8$

This parabola has the vertex at point (6,8), opens upwards, therefore does not intersect the x-axis.

The graph of this parabola is 1.

D. $y=-(x+8)(x-6)$

This parabola has two x-intercepts at points (6,0) and (-8,0).

The leading coefficient is -1 > 0, then the parabola opens downwards.

The only possible choice is parabola 2.

4 0

3 years ago

Read 2 more answers