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3 years ago

10

Help please ! I’m stuck

1 answer:

Troyanec [42]3 years ago

4 0

Y=1x-10 (sorry I could be wrong)

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What is the area of the triangle

Anna11 [10]

Answer:

10

Step-by-step explanation:

1. The area of triangle is = (A*h)/2, where h is the height grounded on side A.

2. Given the height 5 and the side 4 we get S(area)=(5*4)/2=10.

4 0

3 years ago

show me how to work this problem, point A(7,9)is dilated from the origin by scale factor r was= 6. what are the coordinates of p

frutty [35]

If the point is dilated from the origin by a scale factor of 6

Then the coordinates of point A becomes

A( 6x7 , 6x9)

$A_{0,6}(7,9)=(42,\text{ 54)}$

6 0

1 year ago

PLEASE HELP<br><br> Log,3 7=x is equivalent to which of the following (choose all correct answers)

ELEN [110]

Answer:

should be 3^X = 7 and X = log7/log3

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

In circle P with MLNRQ =60, find the mLNPQ

Diano4ka-milaya [45]

Answer is 120
AngNPQ=2*angNRQ=120

7 0

3 years ago

The 5th term in a geometric sequence is 160. The 7th term is 40. What are possible values of the 6th term of the sequence?

omeli [17]

Answer:

C. The 6th term is positive/negative 80

Step-by-step explanation:

Given

Geometric Progression

$T_5 = 160$

$T_7 = 40$

Required

$T_6$

To get the 6th term of the progression, first we need to solve for the first term and the common ratio of the progression;

To solve the common ratio;

Divide the 7th term by the 5th term; This gives

$\frac{T_7}{T_5} = \frac{40}{160}$

Divide the numerator and the denominator of the fraction by 40

$\frac{T_7}{T_5} = \frac{1}{4}$ ----- equation 1

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Where n is the nth term

So,

$T_7 = a r^{6}$

$T_5 = a r^{4}$

Substitute the above expression in equation 1

$\frac{T_7}{T_5} = \frac{1}{4}$ becomes

$\frac{ar^6}{ar^4} = \frac{1}{4}$

$r^2 = \frac{1}{4}$

Square root both sides

$r = \sqrt{\frac{1}{4}}$

r = ± $\frac{1}{2}$

Next, is to solve for the first term;

Using $T_5 = a r^{4}$

By substituting 160 for T5 and ± $\frac{1}{2}$ for r;

We get

$160 = a \frac{1}{2}^{4}$

$160 = a \frac{1}{16}$

Multiply through by 16

$16 * 160 = a \frac{1}{16} * 16$

$16 * 160 = a$

$2560 = a$

Now, we can easily solve for the 6th term

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Here, n = 6;

$T_6 = a r^{6-1}$

$T_6 = a r^5$

$T_6 = 2560 r^5$

r = ± $\frac{1}{2}$

So,

$T_6 = 2560( \frac{1}{2}^5)$ or $T_6 = 2560( \frac{-1}{2}^5)$

$T_6 = 2560( \frac{1}{32})$ or $T_6 = 2560( \frac{-1}{32})$

$T_6 = 80$ or $T_6 = -80$

$T_6 =$ ±80

Hence, the 6th term is positive/negative 80

8 0

3 years ago