What expression is not equivalent to this expression: 75a + 50b

Amanda receives money from her relatives every year on her birthday. Last year, she

umka21 [38]

Answer:

We can call last years amount as 100 percent. that means 68-100. We want to know was 85 is so we can say 85-x. We have 68-100

85-x

Cross multiplying and solving for x we get 8500/68 =x x=125, so the answer is 125.

8 0

3 years ago

I am a two digit number above 50. the sum of my digits is twelve and the difference of my digits is 4. who am I?

Ne4ueva [31]

Answer:

The answer should be 84

Step-by-step explanation:

12 - 4 = 8

8 + 4 = 12

8 - 4 = 4

Hopefully this helps :3 sorry if wrong :( plz mark brainiest if correct :D Your bootiful/handsome! Have a great day luv <3

-Bee~

7 0

3 years ago

PART A: A landmark on the first map is a triangle with side lengths of 3 cm, 4 cm, and 5 cm. What are the side lengths of the tr

Greeley [361]

Complete Question:

Johnny printed two maps of a walking trail near his home. The length of the walking trail on the first map is 8 cm.

(a) Choose a length between 5 cm and 15 cm for the walking trail on the second map: ________cm.

(b) Determine the scale factor from the first map to the second map.

(c) A landmark on the first map is a triangle with side lengths of 3 cm, 4 cm, and 5 cm. What are the side lengths of the triangle landmark on the second map?

(d) Draw one of the triangles from part C. Label the side lengths and vertices accordingly. You may use this drawing tool or draw your triangle on paper:

Answer:

(a) Length = 4cm

(b) Scale factor = 0.5

(c) 3cm, 4cm and 5cm are represented by 1.5cm, 2cm and 2.5cm respectively, on the second scale

(d) See attachment for triangle

Step-by-step explanation:

(a) Choose a length between 5 cm and 15 cm

$Length =4cm$ --- This is solely up to you (you can make use of any length between 5 cm and 15 cm)

(b) The scale factor

The scale factor (k) is calculated as:

$k = \frac{New\ Length}{Old\ Length}$

$k = \frac{4cm}{8cm}\\$

$k = 0.5$

(c) What are side lengths of 3 cm, 4 cm, and 5 cm on the second landmark

Using:

$k = \frac{New\ Length}{Old\ Length}$

The old lengths, in this case are: 3cm, 4cm and 5cm

Make New length the subject

$New\ Length = k * Old\ Length$

When $Length = 3cm$

$New\ Length = 0.5 * 3cm = 1.5cm$

When $Length = 4cm$

$New\ Length = 0.5 * 4cm = 2cm$

When $Length = 5cm$

$New\ Length = 0.5 * 5cm = 2.5cm$

So: 3cm, 4cm and 5cm are represented by 1.5cm, 2cm and 2.5cm respectively, on the second scale

(d) See attachment for triangle

7 0

3 years ago

Prove that a chord is parallel to the tangent drawn at the midpoint of its arc

galina1969 [7]

Need more information

7 0

3 years ago

Read 2 more answers

X-y+z=-4 3x+2y-z=5 -2x+3y-z=15 How do I solve this?

Lera25 [3.4K]

$1)\ \ \ x-y+z=-4\ \ \ \Rightarrow\ \ \ z=-4-x+y\\\\2)\ \ \ 3x+2y-z=5 \\.\ \ \ \ \Rightarrow\ \ 3x+2y-(-4-x+y)=5 \\.\ \ \ \ \Rightarrow\ \ 3x+2y+4+x-y=5 \\.\ \ \ \ \Rightarrow\ \ 4x+y=1\\\\3)\ \ \ -2x+3y-z=15\\.\ \ \ \ \Rightarrow\ \ -2x+3y-(-4-x+y)=15\\.\ \ \ \ \Rightarrow\ \ -2x+3y+4+x-y=15\\.\ \ \ \ \Rightarrow\ \ -x+2y=11\\--------------------\\$

$z=-4-x+y\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ and\\ \left \{ {{4x+y=1\ \ \ \ } \atop {-x+2y=11\ /\cdot4}} \right. \\\\\left \{ {{4x+y=1\ \ \ \ } \atop {-4x+8y=44}} \right. \\-------\\y+8y=1+44\\9y=45\ /:9\\y=5\\\\-x+2y=11\ \ \ \Rightarrow\ \ \ x=2y-11\ \ \ \Rightarrow\ \ \ x=2\cdot5-11=-1\\\\z=-4-x+y=-4-(-1)+5=-4+1+5=2\\\\Ans.\ x=-1\ \ \ and\ \ \ y=5\ \ \ and\ \ \ z=2$

5 0

3 years ago