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3 years ago

7 th grade algebra, needing help to solve. 11c-8+5c-8

Mathematics

2 answers:

Lesechka [4]3 years ago

7 0

Answer:

16c-16

Step-by-step explanation:

11c-8+5c-8

16c -8 -8

16c-16

melamori03 [73]3 years ago

6 0

16c-16 hope this helps

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Complete the identity.<br> 1) sec^4 x + sec^2 x tan^2 x - 2 tan^4 x = ?

Alecsey [184]

Answer:

See Explanation

Step-by-step explanation:

<em>Question like this are better answered if there are list of options; However, I'll simplify as far as the expression can be simplified</em>

Given

$sec^4 x + sec^2 x tan^2 x - 2 tan^4 x$

Required

Simplify

$(sec^2 x)^2 + sec^2 x tan^2 x - 2 (tan^2 x)^2$

Represent $sec^2x$ with a

Represent $tan^2x$ with b

The expression becomes

$a^2 + ab- 2 b^2$

Factorize

$a^2 + 2ab -ab- 2 b^2$

$a(a + 2b) -b(a+ 2 b)$

$(a -b) (a+ 2 b)$

Recall that

$a = sec^2x$

$b = tan^2x$

The expression $(a -b) (a+ 2 b)$ becomes

$(sec^2x -tan^2x) (sec^2x+ 2 tan^2x)$

..............................................................................................................................

In trigonometry

$sec^2x =1 +tan^2x$

Subtract $tan^2x$ from both sides

$sec^2x - tan^2x =1 +tan^2x - tan^2x$

$sec^2x - tan^2x =1$

..............................................................................................................................

Substitute 1 for $sec^2x - tan^2x$ in $(sec^2x -tan^2x) (sec^2x+ 2 tan^2x)$

$(1) (sec^2x+ 2 tan^2x)$

Open Bracket

$sec^2x+ 2 tan^2x$ ------------------This is an equivalence

$(secx)^2+ 2 (tanx)^2$

Solving further;

................................................................................................................................

In trigonometry

$secx = \frac{1}{cosx}$

$tanx = \frac{sinx}{cosx}$

Substitute the expressions for secx and tanx

................................................................................................................................

$(secx)^2+ 2 (tanx)^2$ becomes

$(\frac{1}{cosx})^2+ 2 (\frac{sinx}{cosx})^2$

Open bracket

$\frac{1}{cos^2x}+ 2 (\frac{sin^2x}{cos^2x})$

$\frac{1}{cos^2x}+ \frac{2sin^2x}{cos^2x}$

Add Fraction

$\frac{1 + 2sin^2x}{cos^2x}$ ------------------------ This is another equivalence

................................................................................................................................

In trigonometry

$sin^2x + cos^2x= 1$

Make $sin^2x$ the subject of formula

$sin^2x= 1 - cos^2x$

................................................................................................................................

Substitute the expressions for $1 - cos^2x$ for $sin^2x$

$\frac{1 + 2(1 - cos^2x)}{cos^2x}$

Open bracket

$\frac{1 + 2 - 2cos^2x}{cos^2x}$

$\frac{3 - 2cos^2x}{cos^2x}$ ---------------------- This is another equivalence

8 0

3 years ago

The function, f(x)=45/x models the volume of a gas in a balloon under xx units of pressure at a constant temperature. Which best

JulijaS [17]

My guess will be postive real numbers since pressure can not be negative or imaginary but can be a fraction or irrational

3 0

3 years ago

Help with segment relationships in circles...picture attached.

Reika [66]

The answer: (D)

Explorer acc nss as time yyy has a new one in the middle school in the school and is the junior high junior in junior high junior college

6 0

3 years ago

Use the equation and type the ordered-pairs. y=3^x

aleksandrvk [35]

The ordered pairs of the equation y = 3^x are (0,1), (1,3) and (2,9)

<h3>How to type the ordered pairs?</h3>

The equation is given as:

y = 3^x

Let x = 0, 1 and 2

y = 3^0 = 1

y = 3^1 = 3

y = 3^2 = 9

So, we have the following ordered pairs (0,1), (1,3) and (2,9)

Hence, the ordered pairs of the equation y = 3^x are (0,1), (1,3) and (2,9)

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7 0

2 years ago

4. Tests are performed to determine the Tool Life of a high speed steel single edge cutting tool. At a cutting speeds (v) of 180

podryga [215]

Answer:(a) n=0.233

Step-by-step explanation:

Given

We Know according to Taylor tool life

$VT^n=c$

V=cutting speed

T=tool life

n=tool life exponent

c=constant

$180\times \left ( 4.25\right )^n=110\times \left ( 35\right )^n$

$\frac{180}{110}=\left ( \frac{35}{4.25}\right )^n$

$1.636=\left ( 8.235\right )^n$

taking log both side

$\ln \left ( 1.636\right )=n\ln \left ( 8.235\right )$

n=0.233

3 0

3 years ago