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DaniilM [7]
4 years ago
6

Which sentence explains the correct first step in the solution of this equation? 4(3+x)=9

Mathematics
1 answer:
QveST [7]4 years ago
6 0

Answer:

Multiply 4(3+x)

Step-by-step explanation:

First:  4*3=12 so, 12+3x=9

12+3x=9

-12

3x=-3

divide by 3

Solution: x=-1

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Question one pls<br> algebra 2<br> logarithms
nirvana33 [79]

Condense the logarithm on the right side:

3\log_b(p)-\left(2\log_b(t)+\dfrac12\log_b(r)\right)

n\log_b(m)=\log_b(m^n) \implies \log_b(p^3)-\left(\log_b(t^2)+\log_b\left(r^{\frac12}\right)\right)

\log_b(m)+\log(n)=\log_b(mn) \implies \log_b(p^3)-\log_b\left(t^2r^{\frac12}\right)

\log_b(m)-\log(n)=\log_b\left(\dfrac mn\right) \implies \log_b\left(\dfrac{p^3}{t^2r^{\frac12}}\right)

So,

x=\dfrac{p^3}{t^2r^{\frac12}}

and r^{\frac12}=\sqrt{r}, which makes (4) the correct choice.

4 0
3 years ago
Find the area of quadrilateral ABCD whose vertices are A(1,1) B(7,-3) C(12,2) D(7,21)
sasho [114]

Answer:

The area of quadrilateral ABCD is 139 unit^2.

Step-by-step explanation:

Given:

Quadrilateral ABCD whose vertices are A(1,1) B(7,-3) C(12,2) D(7,21).

Now, to find the area.

The coordinates of the quadrilateral are A(1,1), B(7,-3), C(12,2), D(7,21).

So, to find the area we need to bisect the quadrilateral ABCD and get the triangles ABC and ADC and then calculate their areas:

In Δ ABC:

A(x_1,y_1)=(1,1)\:,\:B(x_2,y_2)=(7,-3)\:and\:C(x_3,y_3)=(12,2)

Now, to get the area of triangle ABC:

Area\,of\,triangle\,=\,\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|

Area\,of\,triangle\,=\,\frac{1}{2}\left|1(-3-2)+(7)(2-1)+12(1--3)\right|

Area\,of\,triangle\,=\,\frac{1}{2}\left|1(-5)+(7)(1)+12(4)\right|

On solving we get:

Area\,of\,triangle\,=25.

In Δ ADC:

A(x_1,y_1)=(1,1)\:,\:D(x_2,y_2)=(7,21)\:and\:C(x_3,y_3)=(12,2)

Now, to get the area of triangle ADC:

Area\,of\,triangle\,=\,\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|

Area\,of\,triangle\,=\,\frac{1}{2}\left|1(21-2)+(7)(2-1)+12(1-21)\right|

On solving it by same process as above we get:

Area\,of\,triangle\,=114

Now, to get the area of the quadrilateral we add the areas of the triangles ABC and ADC:

25+114\\=25+114\\=139\ unit^2

Therefore, area of quadrilateral ABCD is 139 unit^2.

4 0
3 years ago
What is the value of 11−3m , when m = 2? 5 6 10 12
Leona [35]
11 - 3m = 11 - 3(2) = 11 - 6 = 5
3 0
3 years ago
What is the smaller angle of (x+20) (5x+10)
Lera25 [3.4K]

Answer:

Step-by-step explanation:

4 0
2 years ago
The equivalent of 6/11 is a repeating decimal.How many digits are in the repeating decimal?
JulijaS [17]
Hi there!

There are infinite digits in a repeating decimaI. 

Usually, repeating decimals go on forever. They rarely stop ;)

Hope this helps!
8 0
4 years ago
Read 2 more answers
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