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3 years ago

The perimeter of a square garden is 128m. Find the area of the garden in hectares?five star

Mathematics

1 answer:

aleksandrvk [35]3 years ago

7 0

Answer:

1024m

Step-by-step explanation:

128÷4=32

32×32=1024

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Which ratio does not belong with 2/6<br> a) 1/3 b) 3/9 c)6/15 d) 12/36

just olya [345]

Answer:

C)6/15

Step-by-step explanation:

it reduces to 2/5 no 1/3

12/36 --> 1/3

8 0

3 years ago

PLEASE HELP!!!!!!!!!

attashe74 [19]

Math homework, im excillent at math but dm me [email protected] and ill sent you the A

4 0

3 years ago

Suppose θ is an angle in the standard position whose terminal side is in Quadrant IV and cot θ= -6/7 . Find the exact values of

Llana [10]

Answer:

Part 1) $csc(\theta)=-\frac{\sqrt{85}}{7}$

Part 2) $sin(\theta)=-\frac{7}{\sqrt{85}}$ or $sin(\theta)=-\frac{7\sqrt{85}}{85}$

Part 3) $tan(\theta)=-\frac{7}{6}$

Part 4) $cos(\theta)=\frac{6}{\sqrt{85}}$ or $cos(\theta)=\frac{6\sqrt{85}}{85}$

Part 5) $sec(\theta)=\frac{\sqrt{85}}{6}$

Step-by-step explanation:

we know that

The angle theta lie on the IV Quadrant

sin(θ) is negative

cos(θ) is positive

tan(θ) is negative

sec(θ) is positive

csc(θ) is negative

step 1

Find the value of csc(θ)

we know that

$1+cot^{2}(\theta)=csc^{2}(\theta)$

we have

$cot(\theta)=-\frac{6}{7}$

substitute

$1+(-\frac{6}{7})^{2}=csc^{2}(\theta)$

$1+\frac{36}{49}=csc^{2}(\theta)$

$\frac{85}{49}=csc^{2}(\theta)$ rewrite

$csc(\theta)=-\frac{\sqrt{85}}{7}$ ----> remember that is negative

step 2

Find the value of sin(θ)

we know that

$csc(\theta)=\frac{1}{sin(\theta)}$

we have

$csc(\theta)=-\frac{\sqrt{85}}{7}$

therefore

$sin(\theta)=-\frac{7}{\sqrt{85}}$

$sin(\theta)=-\frac{7\sqrt{85}}{85}$

step 3

Find the value of tan(θ)

we know that

$tan(\theta)=\frac{1}{cot(\theta)}$

we have

$cot(\theta)=-\frac{6}{7}$

therefore

$tan(\theta)=-\frac{7}{6}$

step 4

Find the value of cos(θ)

we know that

$sin^{2}(\theta)+cos^{2}(\theta)=1$

we have

$sin(\theta)=-\frac{7}{\sqrt{85}}$

substitute

$(-\frac{7}{\sqrt{85}})^{2}+cos^{2}(\theta)=1$

$\frac{49}{85}+cos^{2}(\theta)=1$

$cos^{2}(\theta)=1-\frac{49}{85}$

$cos^{2}(\theta)=\frac{36}{85}$

$cos(\theta)=\frac{6}{\sqrt{85}}$ ------> the cosine is positive

$cos(\theta)=\frac{6\sqrt{85}}{85}$

step 5

Find the value of sec(θ)

we know that

$sec(\theta)=\frac{1}{cos(\theta)}$

we have

$cos(\theta)=\frac{6}{\sqrt{85}}$

therefore

$sec(\theta)=\frac{\sqrt{85}}{6}$ ----> is positive

8 0

4 years ago

Could someone pls help me out ???

professor190 [17]

Answer:

Option C.

Step-by-step explanation:

$\sf\:x^{2}-3x+7=0$

Use the biquadratic formula.

$\sf\:\x=\frac{-\left(-3\right)(+/-)\sqrt{\left(-3\right)^{2}-4\times 7}}{2}$

$\sf\:x=\frac{-\left(-3\right)(+/-)\sqrt{9-4\times 7}}{2}$

$\sf\:x=\frac{-\left(-3\right)(+/-)\sqrt{9-28}}{2}$

$\sf\:x=\frac{-\left(-3\right)(+/-)\sqrt{-19}}{2}$

$\sf\:x=\frac{3(+/-)\sqrt{19}}{2}$

Positive sign:

$\boxed{\sf\:x=\frac{3+\sqrt{19}}{2}}$

Negative sign:

$\boxed{\sf\:x=\frac{3-\sqrt{19}}{2}}$

Hope it helps ⚜

5 0

2 years ago

A health and fitness club surveys 40 randomly selected members and found that the average weight of those questioned is 157 lb.

svp [43]

Answer:

B. Statistic

Step-by-step explanation:

The average weight of those questioned which is 157 lb. from the survey of 40 randomly selected member of a health and fitness club is an example of a statistic.

It is a calculated numerical value that characterises some aspects of a sample set of data.

Statistic refers to a numerical piece of information that uses quantified models, representations and synopses for a given set of a study. Statistic is used to to estimate the true value of a parameter in a population.

6 0

3 years ago