1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Ket [755]
3 years ago
15

Help me out?? thank you so much whoever does!!

Mathematics
1 answer:
nasty-shy [4]3 years ago
6 0

43º é  th

dgggg  ggddddddd

You might be interested in
Please show work on how to do this​
Irina-Kira [14]

Answer:

  • - 2.71

Step-by-step explanation:

  • 2· 4^{v + 4} + 20 = 32
  • 2· 4^{v + 4} = 12
  • 4^{v + 4} = 6
  • (v + 4) log 4 = log 6
  • v + 4 = log 6 / log 4
  • v + 4 = 1.29  (<em>rounded</em>)
  • v = 1.29 - 4
  • v = - 2.71
5 0
3 years ago
Solve for x. <br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%7D%7B2%7D%20%20%3D%20%20-%207" id="TexFormula1" title=" \frac{
weeeeeb [17]

Answer:

x = -14

Step-by-step explanation:

\frac{x}{2} = -7

multiply 2 on both sides

the 2's cancel out on the fraction side, leaving x = -7(2)

multiply -7 and 2 to get -14

x = -14

4 0
3 years ago
Read 2 more answers
Philip made a total of 9 bracelets and necklaces from 120 inches of cord. He used 8 inches of cord for each bracelet and 20 inch
Viktor [21]

Answer:

So Philip made 5 bracelets and 4 necklaces.

Step-by-step explanation:

Let x = number of bracelets and y = number of necklaces.

Since we have a total of 9 bracelets and necklaces,

x + y = 9 (1)

Also, we have 8 inches of cord for each bracelet and 20 inches of cord for each necklace, then the total length for the bracelet is 8x and that for the necklace is 20y.

So, the total length for both is 8x + 20y. Since the total length of cord used is 120 inches,

8x + 20y = 120 (2)

Simplifying it we have

2x + 5y = 30  (3).

Writing equations (1) and (3) in matrix form, we have

\left[\begin{array}{ccc}1&1\\2&5\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}9\\30\end{array}\right]

Using Cramer's rule to solve for x and y,

x = det \left[\begin{array}{ccc}9&1\\30&5\end{array}\right] /det \left[\begin{array}{ccc}1&1\\2&5\end{array}\right] \\

x = (9 × 5 - 30 × 1) ÷ (1 × 5 - 1 × 2)

x = (45 - 30) ÷ (5 - 2)

x = 15 ÷ 3

x = 5

y = det \left[\begin{array}{ccc}1&9\\2&30\end{array}\right] /det \left[\begin{array}{ccc}1&1\\2&5\end{array}\right] \\

y = (30 × 1 - 9 × 2) ÷ (1 × 5 - 1 × 2)

y = (30 - 18) ÷ (5 - 2)

y = 12 ÷ 3

y = 4

So Philip made 5 bracelets and 4 necklaces.

3 0
3 years ago
A line tangent to a circle and a radius of the same circle intersect at the point of tangency to create an angle of what measure
aliya0001 [1]

A theorem states that, given a circle with center C and a point P on the circumference, the tangent line through P and the radius CP are perpendicular.

So, the answer is 90 degrees.

3 0
3 years ago
Rewrite using powers of ten 32,000,000
mamaluj [8]
32×10×10×10×10×10×10
5 0
3 years ago
Read 2 more answers
Other questions:
  • Evaluate using the values given.<br> ZX - [z - (4 + x)/6] when x = 2, z = 6
    14·1 answer
  • 1/4 (3s-14)=4 tell me how you solved it
    10·1 answer
  • Interest is calculated by ____.
    11·2 answers
  • Which of the following situations best models the materials needed to build the fence for the warehouse lot shown in the general
    6·1 answer
  • Which could be the first step in solving the equation represented by the model below? 3 x tiles and 4 negative 1 tiles = 2 x til
    7·2 answers
  • Jason enjoys watching the squirrels in his neighborhood park. They eat the red oak acorns. After the city removed 4 diseased red
    11·1 answer
  • 3/7×=33 what is the answer to this math problem
    13·1 answer
  • Someone please help me on this
    7·1 answer
  • Question 2<br> What is the y-intercept in the equation y = 4x - 3?<br> 4<br> 0<br> 3
    7·1 answer
  • A) La suma de dos números es menor que cuatro a+b&lt;4
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!