Find the value of the expression: x+5-6 when x=5 Step by step please

Mathematics

1 answer:

Butoxors [25]3 years ago

8 0

Answer:

Step-by-step explanation:

Since x = 5, you can substitute 5 into the x value of the expression: 5 + 5 - 6. Then, you add the 5 + 5 to get 10 in which you subtract 6 from. The answer is 4.

You might be interested in

Melania has two jobs. During the day she works as an office clerk, and in the evening she works as a cashier. her office job pay

bulgar [2K]

Lets x = # of hours Melania works at office clerk
and y = # of hours Melania works as a cashier

x + y = 38 so x = 38 - y
13x + 9.25y = 434

substitute x = 38 - y into 13x + 9.25y = 434

13x + 9.25y = 434
13(38 - y) + 9.25y = 434
494 - 13y + 9.25y = 434
-3.75y = -60
y = 16
x = 38 - y so x = 38 - 16 = 22

answer
Melania works at office clerk = 22 hours
Melania works as a cashier = 16 hours

7 0

3 years ago

Help me pls don’t put random

Vladimir [108]

I think it’s A
Sorry if I’m wrong

4 0

3 years ago

Read 2 more answers

7 divided by 1/2 <img src="https://tex.z-dn.net/?f=7%20%5Cdiv%7C%201" id="TexFormula1" title="7 \div| 1" alt="7 \div| 1" alig

Elan Coil [88]

The answer should be 14

7 0

2 years ago

Read 2 more answers

What is the following product? ^3 square root 24 times ^3 square root 45

torisob [31]

Answer:

$6 \sqrt[3]{5}$

Step-by-step explanation:

For the problem, $\sqrt[3]{24} *\sqrt[3]{45}$ , use rules for simplifying cube roots. Under the operations of multiplication and division, if the roots have the same index (here it is 3) you can combine them.

$\sqrt[3]{24} *\sqrt[3]{45} = \sqrt[3]{24*45}$

You can multiply it out completely, however to simplify after you'll need to pull out perfect cubes. Factor 24 and 45 into any perfect cube factors which multiply to each number. If none are there, then prime factors will do. You can group factors together such as 3*3*3 which is 27 and a perfect cube.

$\sqrt[3]{24*45} =\sqrt[3]{3*8*5*3*3} = 6 \sqrt[3]{5}$

7 0

3 years ago

Write the equation of a hyperbola with vertices (0, -4) and (0, 4) and foci (0, -5) and (0, 5).

andre [41]

Check the picture below. So, more or less looks like so.

notice, the center is clearly at the origin, and notice how long the "a" component is, also, bear in mind that, is opening towards the y-axis, that means the fraction with the "y" variable is the positive one.

Also notice, the "c" distance from the center to either foci, is just 5 units.

$\bf \textit{hyperbolas, vertical traverse axis }\\\\
\cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1
\qquad 
\begin{cases}
center\ ({{ h}},{{ k}})\\
vertices\ ({{ h}}, {{ k}}\pm a)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{{{ a }}^2+{{ b }}^2}
\end{cases}\\\\
-------------------------------\\\\$

$\bf \begin{cases}
h=0\\
k=0\\
a=4\\
c=5
\end{cases}\implies \cfrac{(y-{{ 0}})^2}{{{ 4}}^2}-\cfrac{(x-{{ 0}})^2}{{{ b}}^2}=1\implies \cfrac{y^2}{16}-\cfrac{x^2}{b^2}=1
\\\\\\
c=\sqrt{a^2+b^2}\implies c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\\\\\\
\sqrt{5^2-4^2}=b\implies \boxed{3=b}
\\\\\\
\cfrac{y^2}{16}-\cfrac{x^2}{3^2}=1\implies \boxed{\cfrac{y^2}{16}-\cfrac{x^2}{9}=1}$

7 0

4 years ago