Please help me I really need help

Can anyone help me with these i need them befire tomorrow

mash [69]

When you have a scale factor as percentage, is better to convert that percentage into a fraction by divide it by 100%. That way, it will be much easier to scale your figures.

1. The scale factor we need to apply to our rectangle is 60%. Firts, lets convert our scale factor into a fraction: $\frac{60}{100} = \frac{3}{5}$ . Next, we just need to multiply each side of our rectangle by our fraction:
length= $16( \frac{3}{5})=9.6$
width= $8( \frac{3}{5} )=4.8$
So, our new rectangle will have a length of 9.6 units and a width of 4.8 units.

2. This time the scale factor is 120%. Just as before, lets convert it into a fraction: $\frac{120}{100} = \frac{6}{5}$ , and just as before lets multiply each side of our triangle by our fraction:
base= $10( \frac{6}{5} )=12$
side= $8( \frac{6}{5} )=9.6$
So, our new isosceles triangle will have a base of 12 units and tow sides of length 9.6 units each.

3. This time the scale factor is 150%. Just as before, lets convert it into a fraction: $\frac{150}{100} = \frac{3}{2}$ , and just as before lets multiply each side of our polygon by our fraction:
$12( \frac{3}{2} )=18$
$15( \frac{3}{2} )=22.5$
$13( \frac{3}{2} )=19.5$
So, our new polygon will have a side of 12 length units, tow sides of 15 length units, and two side of 15 length units.

4. This time our factor is 75%. Just as before, lets convert it into a fraction: $\frac{75}{100} = \frac{3}{4}$ , and just as before lets multiply each side of our parallelogram by our fraction:
side= $25( \frac{3}{4} )=18.75$
So, our new parallelogram will have tow sides of length 18.75 units.

7 0

4 years ago

1 3/4+1/6+____=7 1/2<br> Plz help

Mademuasel [1]

Answer:

7/4 +1/6 +X=15/2

1/6+X=15/2-7/4

1/6+X=30/4-7/4

X=23/4 - 1/6

X= 138-4/24

134/24

5.58333

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Find the area of the circle whose equation is x2+y2=6x-8y

Semenov [28]

Answer:

Given that the equation of a circle is :

$\green{ \boxed{\boxed{\begin{array}{cc} {x}^{2} + {y}^{2} = 6x - 8y \\ = > {x}^{2} + {y}^{2} - 6x + 8y = 0 \\ = > {x}^{2} + {y}^{2} + 2 \times ( - 3) \times x + 2 \times 4 \times y = 0 \\ \\ \sf \: standard \: equation \: o f \: circle \: is : \\ {x}^{2} + {x}^{2} + 2gx + 2fy + c = 0 \\ \\ \sf \: by \: comparing \\ \\ g = - 3 \\ f = 4 \\ c = 0 \\ \\ \sf \: radius \: \: r = \sqrt{ {g}^{2} + {f}^{2} - c } \\ = \sqrt{ {( - 3)}^{2} + {4}^{2} - 0 } \\ = \sqrt{9 + 16} \\ = \sqrt{25} \\ = 5 \: unit \\ \\ \bf \: area \: = \pi {r}^{2} \\ = \pi \times {5}^{2} \\ =\pink{ 25\pi \: { unit }^{2} }\end{array}}}}$